有效地合并两个排序的数组，并增加O(1)个额外空间

给定两个排序的数组，我们需要在O((n + m)* log(n + m))的时间内将它们与O(1)的额外空间合并到一个排序的数组中，此时n是第一个数组的大小，并且m是第二个数组的大小。

例子：

Input: ar1[] = {10};
       ar2[] = {2, 3};
Output: ar1[] = {2}
        ar2[] = {3, 10}  

Input: ar1[] = {1, 5, 9, 10, 15, 20};
       ar2[] = {2, 3, 8, 13};
Output: ar1[] = {1, 2, 3, 5, 8, 9}
        ar2[] = {10, 13, 15, 20}

我们在下面的文章中讨论了二次时间解决方案。

在这篇文章中，讨论了一个更好的解决方案。
这个想法：我们开始比较彼此相距遥远而不是相邻的元素。
对于每次通过，我们都计算间隙并比较间隙右侧的元素。每次通过，间隙减小到除以2的上限值。

例子：

First example: 
a1[] = {3 27 38 43}, 
a2[] = {9 10 82}
Start with 
gap =  ceiling of n/2 = 4 
[This gap is for whole merged array]
3 27 38 43   9 10 82 
3 27 38 43   9 10 82
3 10 38 43   9 27 82

gap = 2:
3 10 38 43   9 27 82
3 10 38 43   9 27 82
3 10 38 43   9 27 82 
3 10 9 43   38 27 82
3 10 9 27   38 43 82

gap = 1:
3 10 9 27   38 43 82
3 10 9 27   38 43 82
3 9 10 27   38 43 82
3 9 10 27   38 43 82
3 9 10 27   38 43 82
3 9 10 27   38 43 82
Output : 3 9 10 27 38 43 82

Second Example: 
a1[] = {10 27 38 43 82}, 
a2[] = {3 9}
Start with gap = ceiling of n/2 (4):
10 27 38 43 82   3 9 
10 27 38 43 82   3 9
10 3 38 43 82   27 9
10 3 9 43 82   27 38

gap = 2:
10 3 9 43 82   27 38
9 3 10 43 82   27 38
9 3 10 43 82   27 38
9 3 10 43 82   27 38
9 3 10 27 82   43 38
9 3 10 27 38   43 82

gap = 1
9 3 10 27 38   43 82
3 9 10 27 38   43 82
3 9 10 27 38   43 82
3 9 10 27 38   43 82
3 9 10 27 38   43 82
3 9 10 27 38   43 82


Output : 3 9 10 27 38   43 82

下面是上述想法的实现：

C++

// Merging two sorted arrays with O(1)
// extra space
#include 
using namespace std;
 
// Function to find next gap.
int nextGap(int gap)
{
    if (gap <= 1)
        return 0;
    return (gap / 2) + (gap % 2);
}
 
void merge(int* arr1, int* arr2, int n, int m)
{
    int i, j, gap = n + m;
    for (gap = nextGap(gap);
         gap > 0; gap = nextGap(gap))
    {
        // comparing elements in the first array.
        for (i = 0; i + gap < n; i++)
            if (arr1[i] > arr1[i + gap])
                swap(arr1[i], arr1[i + gap]);
 
        // comparing elements in both arrays.
        for (j = gap > n ? gap - n : 0;
             i < n && j < m;
             i++, j++)
            if (arr1[i] > arr2[j])
                swap(arr1[i], arr2[j]);
 
        if (j < m) {
            // comparing elements in the second array.
            for (j = 0; j + gap < m; j++)
                if (arr2[j] > arr2[j + gap])
                    swap(arr2[j], arr2[j + gap]);
        }
    }
}
 
// Driver code
int main()
{
    int a1[] = { 10, 27, 38, 43, 82 };
    int a2[] = { 3, 9 };
    int n = sizeof(a1) / sizeof(int);
    int m = sizeof(a2) / sizeof(int);
 
    // Function Call
    merge(a1, a2, n, m);
 
    printf("First Array: ");
    for (int i = 0; i < n; i++)
        printf("%d ", a1[i]);
 
    printf("\nSecond Array: ");
    for (int i = 0; i < m; i++)
        printf("%d ", a2[i]);
    printf("\n");
    return 0;
}

Java

// Java program for Merging two sorted arrays
// with O(1) extra space
 
public class MergeTwoSortedArrays {
 
    // Function to find next gap.
    private static int nextGap(int gap)
    {
        if (gap <= 1)
            return 0;
        return (gap / 2) + (gap % 2);
    }
 
    private static void merge(int[] arr1,
                              int[] arr2, int n,
                              int m)
    {
        int i, j, gap = n + m;
        for (gap = nextGap(gap); gap > 0;
             gap = nextGap(gap))
        {
            // comparing elements in the first
            // array.
            for (i = 0; i + gap < n; i++)
                if (arr1[i] > arr1[i + gap])
                {
                    int temp = arr1[i];
                    arr1[i] = arr1[i + gap];
                    arr1[i + gap] = temp;
                }
 
            // comparing elements in both arrays.
            for (j = gap > n ? gap - n : 0;
                 i < n && j < m;
                 i++, j++)
                if (arr1[i] > arr2[j])
                {
                    int temp = arr1[i];
                    arr1[i] = arr2[j];
                    arr2[j] = temp;
                }
 
            if (j < m)
            {
                // comparing elements in the
                // second array.
                for (j = 0; j + gap < m; j++)
                    if (arr2[j] > arr2[j + gap])
                    {
                        int temp = arr2[j];
                        arr2[j] = arr2[j + gap];
                        arr2[j + gap] = temp;
                    }
            }
        }
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int[] a1 = { 10, 27, 38, 43, 82 };
        int[] a2 = { 3, 9 };
 
        // Function Call
        merge(a1, a2, a1.length, a2.length);
 
        System.out.print("First Array: ");
        for (int i = 0; i < a1.length; i++) {
            System.out.print(a1[i] + " ");
        }
 
        System.out.println();
 
        System.out.print("Second Array: ");
        for (int i = 0; i < a2.length; i++) {
            System.out.print(a2[i] + " ");
        }
    }
}
 
// This code is contributed by Vinisha Shah

Python3

# Merging two sorted arrays with O(1)
# extra space
 
# Function to find next gap.
 
 
def nextGap(gap):
 
    if (gap <= 1):
        return 0
    return (gap // 2) + (gap % 2)
 
 
def merge(arr1, arr2, n, m):
 
    gap = n + m
    gap = nextGap(gap)
    while gap > 0:
 
        # comparing elements in
        # the first array.
        i = 0
        while i + gap < n:
            if (arr1[i] > arr1[i + gap]):
                arr1[i], arr1[i + gap] = arr1[i + gap], arr1[i]
 
            i += 1
 
        # comparing elements in both arrays.
        j = gap - n if gap > n else 0
        while i < n and j < m:
            if (arr1[i] > arr2[j]):
                arr1[i], arr2[j] = arr2[j], arr1[i]
 
            i += 1
            j += 1
 
        if (j < m):
 
            # comparing elements in the
            # second array.
            j = 0
            while j + gap < m:
                if (arr2[j] > arr2[j + gap]):
                    arr2[j], arr2[j + gap] = arr2[j + gap], arr2[j]
 
                j += 1
 
        gap = nextGap(gap)
 
 
# Driver code
if __name__ == "__main__":
 
    a1 = [10, 27, 38, 43, 82]
    a2 = [3, 9]
    n = len(a1)
    m = len(a2)
 
    # Function Call
    merge(a1, a2, n, m)
 
    print("First Array: ", end="")
    for i in range(n):
        print(a1[i], end=" ")
    print()
 
    print("Second Array: ", end="")
    for i in range(m):
        print(a2[i], end=" ")
    print()
 
# This code is contributed
# by ChitraNayal

C#

// C# program for Merging two sorted arrays
// with O(1) extra space
using System;
 
class GFG {
 
    // Function to find next gap.
    static int nextGap(int gap)
    {
        if (gap <= 1)
            return 0;
        return (gap / 2) + (gap % 2);
    }
 
    private static void merge(int[] arr1, int[] arr2, int n,
                              int m)
    {
        int i, j, gap = n + m;
        for (gap = nextGap(gap); gap > 0;
             gap = nextGap(gap))
        {
            // comparing elements in the first
            // array.
            for (i = 0; i + gap < n; i++)
                if (arr1[i] > arr1[i + gap])
                {
                    int temp = arr1[i];
                    arr1[i] = arr1[i + gap];
                    arr1[i + gap] = temp;
                }
 
            // comparing elements in both arrays.
            for (j = gap > n ? gap - n : 0; i < n && j < m;
                 i++, j++)
                if (arr1[i] > arr2[j])
                {
                    int temp = arr1[i];
                    arr1[i] = arr2[j];
                    arr2[j] = temp;
                }
 
            if (j < m) {
                // comparing elements in the
                // second array.
                for (j = 0; j + gap < m; j++)
                    if (arr2[j] > arr2[j + gap])
                    {
                        int temp = arr2[j];
                        arr2[j] = arr2[j + gap];
                        arr2[j + gap] = temp;
                    }
            }
        }
    }
 
    // Driver code
    public static void Main()
    {
        int[] a1 = { 10, 27, 38, 43, 82 };
        int[] a2 = { 3, 9 };
 
        // Function Call
        merge(a1, a2, a1.Length, a2.Length);
 
        Console.Write("First Array: ");
        for (int i = 0; i < a1.Length; i++) {
            Console.Write(a1[i] + " ");
        }
 
        Console.WriteLine();
 
        Console.Write("Second Array: ");
        for (int i = 0; i < a2.Length; i++) {
            Console.Write(a2[i] + " ");
        }
    }
}
 
// This code is contributed by Sam007.

PHP

 0; $gap = nextGap($gap))
    {
        // comparing elements
        // in the first array.
        for ($i = 0; $i + $gap < $n; $i++)
            if ($arr1[$i] > $arr1[$i + $gap])
            {
                $tmp = $arr1[$i];
                $arr1[$i] = $arr1[$i + $gap];
                $arr1[$i + $gap] = $tmp;
            }
 
        // comparing elements
        // in both arrays.
        for ($j = $gap > $n ? $gap - $n : 0 ;
             $i < $n && $j < $m; $i++, $j++)
            if ($arr1[$i] > $arr2[$j])
            {
                $tmp = $arr1[$i];
                $arr1[$i] = $arr2[$j];
                $arr2[$j] = $tmp;
            }
 
        if ($j < $m)
        {
            // comparing elements in
            // the second array.
            for ($j = 0; $j + $gap < $m; $j++)
                if ($arr2[$j] > $arr2[$j + $gap])
                {
                    $tmp = $arr2[$j];
                    $arr2[$j] = $arr2[$j + $gap];
                    $arr2[$j + $gap] = $tmp;
                }
        }
    }
     
    echo "First Array: ";
    for ($i = 0; $i < $n; $i++)
        echo $arr1[$i]." ";
 
    echo "\nSecond Array: ";
    for ($i = 0; $i < $m; $i++)
        echo $arr2[$i] . " ";
    echo"\n";
 
}
 
// Driver code
$a1 = array(10, 27, 38, 43 ,82);
$a2 = array(3,9);
$n = sizeof($a1);
$m = sizeof($a2);
 
// Function Call
merge($a1, $a2, $n, $m);
 
// This code is contributed
// by mits.
?>

输出

First Array: 3 9 10 27 38 
Second Array: 43 82

O(m + n)时间复杂度的另一种方法：

在这里，我们使用以下技术：

Suppose we have a number A and we want to 
convert it to a number B and there is also a 
constraint that we can recover number A any 
time without using other variable.To achieve 
this we chose a number N which is greater 
than both numbers and add B*N in A.
so A --> A+B*N

To get number B out of (A+B*N) 
we devide (A+B*N) by N (A+B*N)/N = B.

To get number A out of (A+B*N) 
we take modulo with N (A+B*N)%N = A.

-> In short by taking modulo 
we get old number back and taking divide 
we new number.

我们首先找到两个数组的最大元素，然后将其增加一二，以避免在模运算过程中0和最大元素发生冲突。这个想法是从同时开始遍历两个数组。假设a中的元素是a [i]，b中的元素是b [j]，k是下一个最小数将到达的位置。现在，如果k

下面是上述想法的实现：

C++

#include 
using namespace std;
 
void mergeArray(int a[], int b[], int n, int m)
{
    int mx = 0;
     
    // Find maximum element of both array
    for (int i = 0; i < n; i++) {
        mx = max(mx, a[i]);
    }
    for (int i = 0; i < m; i++) {
        mx = max(mx, b[i]);
    }
     
    // increment one two avoid collision of 0 and maximum
    // element of array in modulo operation
    mx++;
    int i = 0, j = 0, k = 0;
    while (i < n && j < m && k < (n + m)) {
         
        // recover back original element to compare
        int e1 = a[i] % mx;
        int e2 = b[j] % mx;
        if (e1 <= e2) {
             
            // update element by adding multiplication
            // with new number
            if (k < n)
                a[k] += (e1 * mx);
            else
                b[k - n] += (e1 * mx);
            i++;
            k++;
        }
        else {
             
            // update element by adding multiplication
            // with new number
            if (k < n)
                a[k] += (e2 * mx);
            else
                b[k - n] += (e2 * mx);
            j++;
            k++;
        }
    }
     
    // process those elements which are left in array a
    while (i < n) {
        int el = a[i] % mx;
        if (k < n)
            a[k] += (el * mx);
        else
            b[k - n] += (el * mx);
        i++;
        k++;
    }
     
    // process those elements which are left in array a
    while (j < m) {
        int el = b[j] % mx;
        if (k < n)
            b[k] += (el * mx);
        else
            b[k - n] += (el * mx);
        j++;
        k++;
    }
     
    // finally update elements by dividing
    // with maximum element
    for (int i = 0; i < n; i++)
        a[i] = a[i] / mx;
 
    // finally update elements by dividing
    // with maximum element
    for (int i = 0; i < m; i++)
        b[i] = b[i] / mx;
 
    return;
}
 
// Driver Code
int main()
{
    int a[] = { 3, 5, 6, 8, 12 };
    int b[] = { 1, 4, 9, 13 };
    int n = sizeof(a) / sizeof(int); // Length of a
    int m = sizeof(b) / sizeof(int); // length of b
 
    // Function Call
    mergeArray(a, b, n, m);
 
    cout << "First array : ";
    for (int i = 0; i < n; i++)
        cout << a[i] << " ";
    cout << endl;
 
    cout << "Second array : ";
    for (int i = 0; i < m; i++)
        cout << b[i] << " ";
    cout << endl;
 
    return 0;
}

Java

import java.io.*;
 
class GFG{
 
static void mergeArray(int a[], int b[],
                       int n, int m)
{
    int mx = 0;
 
    // Find maximum element of both array
    for(int i = 0; i < n; i++)
    {
        mx = Math.max(mx, a[i]);
    }
     
    for(int i = 0; i < m; i++)
    {
        mx = Math.max(mx, b[i]);
    }
     
    // Increment one two avoid collision of
    // 0 and maximum element of array in
    // modulo operation
    mx++;
    int i = 0, j = 0, k = 0;
     
    while (i < n && j < m && k < (n + m))
    {
         
        // Recover back original element
        // to compare
        int e1 = a[i] % mx;
        int e2 = b[j] % mx;
         
        if (e1 <= e2)
        {
             
            // Update element by adding
            // multiplication with new number
            if (k < n)
                a[k] += (e1 * mx);
            else
                b[k - n] += (e1 * mx);
                 
            i++;
            k++;
        }
        else
        {
             
            // Update element by adding
            // multiplication with new number
            if (k < n)
                a[k] += (e2 * mx);
            else
                b[k - n] += (e2 * mx);
                 
            j++;
            k++;
        }
    }
 
    // Process those elements which are
    // left in array a
    while (i < n)
    {
        int el = a[i] % mx;
        if (k < n)
            a[k] += (el * mx);
        else
            b[k - n] += (el * mx);
             
        i++;
        k++;
    }
 
    // Process those elements which are
    // left in array a
    while (j < m)
    {
        int el = b[j] % mx;
         
        if (k < n)
            b[k] += (el * mx);
        else
            b[k - n] += (el * mx);
             
        j++;
        k++;
    }
 
    // Finally update elements by dividing
    // with maximum element
    for(int in = 0; in < n; in++)
        a[in] = a[in] / mx;
 
    // Finally update elements by dividing
    // with maximum element
    for(int in = 0; in < m; in++)
        b[in] = b[in] / mx;
 
    return;
}
 
// Driver code
public static void main(String[] args)
{
    int a[] = { 3, 5, 6, 8, 12 };
    int b[] = { 1, 4, 9, 13 };
     
    // Length of a
    int n = a.length;
     
    // length of b
    int m = b.length;
 
    // Function Call
    mergeArray(a, b, n, m);
 
    System.out.print("First array : ");
    for(int i = 0; i < n; i++)
        System.out.print(a[i] + " ");
         
    System.out.println();
 
      System.out.print("Second array : ");
    for(int i = 0; i < m; i++)
        System.out.print(b[i] + " ");
         
    System.out.println();
}
}
 
// This code is contributed by yashbeersingh42

C＃

using System;
 
public class GFG{
 
  static void mergeArray(int[] a, int[] b,
                         int n, int m)
  {
    int mx = 0;
 
    // Find maximum element of both array
    for(int I = 0; I < n; I++)
    {
      mx = Math.Max(mx, a[I]);
    }
 
    for(int I = 0; I < m; I++)
    {
      mx = Math.Max(mx, b[I]);
    }
 
    // Increment one two avoid collision of
    // 0 and maximum element of array in
    // modulo operation
    mx++;
    int i = 0, j = 0, k = 0;
 
    while (i < n && j < m && k < (n + m))
    {
 
      // Recover back original element
      // to compare
      int e1 = a[i] % mx;
      int e2 = b[j] % mx;
 
      if (e1 <= e2)
      {
 
        // Update element by adding
        // multiplication with new number
        if (k < n)
          a[k] += (e1 * mx);
        else
          b[k - n] += (e1 * mx);
 
        i++;
        k++;
      }
      else
      {
 
        // Update element by adding
        // multiplication with new number
        if (k < n)
          a[k] += (e2 * mx);
        else
          b[k - n] += (e2 * mx);
 
        j++;
        k++;
      }
    }
 
    // Process those elements which are
    // left in array a
    while (i < n)
    {
      int el = a[i] % mx;
      if (k < n)
        a[k] += (el * mx);
      else
        b[k - n] += (el * mx);
 
      i++;
      k++;
    }
 
    // Process those elements which are
    // left in array a
    while (j < m)
    {
      int el = b[j] % mx;
 
      if (k < n)
        b[k] += (el * mx);
      else
        b[k - n] += (el * mx);
 
      j++;
      k++;
    }
 
    // Finally update elements by dividing
    // with maximum element
    for(int In = 0; In < n; In++)
      a[In] = a[In] / mx;
 
    // Finally update elements by dividing
    // with maximum element
    for(int In = 0; In < m; In++)
      b[In] = b[In] / mx;
 
    return;
  }
 
  // Driver code
  static public void Main (){
    int[] a = { 3, 5, 6, 8, 12 };
    int[] b = { 1, 4, 9, 13 };
 
    // Length of a
    int n = a.Length;
 
    // length of b
    int m = b.Length;
 
    // Function Call
    mergeArray(a, b, n, m);
 
    Console.Write("First array : ");
    for(int i = 0; i < n; i++)
      Console.Write(a[i] + " ");
 
    Console.WriteLine();
 
    Console.Write("Second array : ");
    for(int i = 0; i < m; i++)
      Console.Write(b[i] + " ");
 
    Console.WriteLine();
  }
}
 
// This code is contributed by avanitrachhadiya2155

输出

First array : 1 3 4 5 6 
Second array : 8 9 12 13

时间复杂度： O(m + n)

我们在下面的文章中讨论了一种简单有效的解决方案。

在这篇文章中，讨论了一个更好，更简单的解决方案。

这个想法是：我们将遍历第一个数组并将其与第二个数组的第一个元素进行比较。如果第二个数组的第一个元素小于第一个数组，那么我们将交换然后对第二个数组进行排序。

1. 首先，我们必须遍历array1，然后将其与array2的第一个元素进行比较。
   如果它小于array1，则交换我们两个交换。
2. 交换之后，我们将再次对array2进行排序，以便array2的最小元素
   位于第一个位置，我们可以再次与array1交换
3. 为了对array2进行排序，我们将array2的第一个元素存储在变量中，然后将所有元素左移并存储
   array2中的第一个元素，最后一个。

C++

#include 
using namespace std;
 
void mergeArray(int arr1[], int arr2[],
                int n, int m)
{
     
    // Now traverse the array1 and if
    // arr2 first element
    // is less than arr1 then swap
    for(int i = 0; i < n; i++)
    {
        if (arr1[i] > arr2[0])
        {
             
            // Swap
            int temp = arr1[i];
            arr1[i] = arr2[0];
            arr2[0] = temp;
 
            // After swaping we have to sort the array2
            // again so that it can be again swap with
            // arr1
 
            // We will store the firstElement of array2
            // and left shift all the element and store
            // the firstElement in arr2[k-1]
            int firstElement = arr2[0];
 
            int k;
             
            for(k = 1;
                k < m && arr2[k] < firstElement;
                k++)
            {
                arr2[k - 1] = arr2[k];
            }
            arr2[k - 1] = firstElement;
        }
    }
     
    // Read the arr1
    for(int i = 0; i < n; i++)
    {
        cout << arr1[i] << " ";
    }
    cout << endl;
 
    // Read the arr2
    for(int i = 0; i < m; i++)
    {
        cout << arr2[i] << " ";
    }
}
 
// Driver Code
int main()
{
    int arr1[] = { 1, 3, 5, 7 };
    int arr2[] = { 0, 2, 6, 8, 9 };
    int n = 4, m = 5;
     
    mergeArray(arr1, arr2, n, m);
}
 
// This code is contributed by yashbeersingh42

Java

import java.io.*;
 
class GFG {
 
    public static void mergeArray(int[] arr1, int[] arr2)
    {
 
        // length of first arr1
        int n = arr1.length;
 
        // length of second arr2
        int m = arr2.length;
 
        // Now traverse the array1 and if
        // arr2 first element
        // is less than arr1 then swap
        for (int i = 0; i < n; i++) {
 
            if (arr1[i] > arr2[0]) {
 
                // swap
                int temp = arr1[i];
                arr1[i] = arr2[0];
                arr2[0] = temp;
 
                // after swaping we have to sort the array2
                // again so that it can be again swap with
                // arr1
 
                // we will store the firstElement of array2
                // and left shift all the element and store
                // the firstElement in arr2[k-1]
 
                int firstElement = arr2[0];
 
                int k;
                for (k = 1;
                     k < m && arr2[k] < firstElement;
                     k++)
                {
                    arr2[k - 1] = arr2[k];
                }
                arr2[k - 1] = firstElement;
            }
        }
 
        // read the arr1
        for (int i : arr1) {
            System.out.print(i + " ");
        }
 
        System.out.println();
 
        // read the arr2
        for (int i : arr2) {
            System.out.print(i + " ");
        }
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int[] arr1 = { 1, 3, 5, 7 };
 
        int[] arr2 = { 0, 2, 6, 8, 9 };
        mergeArray(arr1, arr2);
    }
}

C＃

using System;
 
class GFG{
     
public static void mergeArray(int[] arr1,
                              int[] arr2)
{
     
    // Length of first arr1
    int n = arr1.Length;
 
    // Length of second arr2
    int m = arr2.Length;
 
    // Now traverse the array1 and if
    // arr2 first element
    // is less than arr1 then swap
    for(int i = 0; i < n; i++)
    {
        if (arr1[i] > arr2[0])
        {
             
            // Swap
            int temp = arr1[i];
            arr1[i] = arr2[0];
            arr2[0] = temp;
 
            // After swaping we have to sort the array2
            // again so that it can be again swap with
            // arr1
 
            // We will store the firstElement of array2
            // and left shift all the element and store
            // the firstElement in arr2[k-1]
            int firstElement = arr2[0];
 
            int k;
            for(k = 1;
                k < m && arr2[k] < firstElement;
                k++)
            {
                arr2[k - 1] = arr2[k];
            }
            arr2[k - 1] = firstElement;
        }
    }
 
    // Read the arr1
    foreach (int i in arr1)
    {
        Console.Write(i + " ");
    }
 
    Console.WriteLine();
 
    // Read the arr2
    foreach(int i in arr2)
    {
        Console.Write(i + " ");
    }
}
 
// Driver Code
static public void Main()
{
    int[] arr1 = { 1, 3, 5, 7 };
    int[] arr2 = { 0, 2, 6, 8, 9 };
     
    mergeArray(arr1, arr2);
}
}
 
// This code is contributed by rag2127

输出

0 1 2 3 
5 6 7 8 9

时间复杂度： O(n * m)
空间复杂度： O(1)