从重复数组中查找丢失的元素

给定两个彼此重复的数组，除了一个元素之外，即缺少一个数组中的一个元素，我们需要找到那个缺失的元素。
例子：

Input:  arr1[] = {1, 4, 5, 7, 9}
        arr2[] = {4, 5, 7, 9}
Output: 1
1 is missing from second array.

Input: arr1[] = {2, 3, 4, 5}
       arr2[] = {2, 3, 4, 5, 6}
Output: 6
6 is missing from first array.

一个简单的解决方案是遍历数组并逐个元素检查元素，并在找到不匹配的元素时标记丢失的元素，但此解决方案需要数组的线性时间过大。
另一种有效的解决方案是基于二进制搜索方法。算法步骤如下：

1. 在更大的数组中开始二分查找并得到 mid 为 (lo + hi) / 2
2. 如果两个数组的值相同，则缺少的元素必须在正确的部分，因此将 lo 设置为 mid
3. 否则将 hi 设置为 mid，因为如果中间元素不相等，则丢失的元素必须位于较大数组的左侧。
4. 对于单个元素和零元素数组，单独处理特殊情况，单个元素本身将是缺失元素。
   如果第一个元素本身不相等，则该元素将是缺失的元素。/li>

下面是上述步骤的实现

C++

// C++ program to find missing element from same
// arrays (except one missing element)
#include 
using namespace std;
 
// Function to find missing element based on binary
// search approach.  arr1[] is of larger size and
// N is size of it.  arr1[] and arr2[] are assumed
// to be in same order.
int findMissingUtil(int arr1[], int arr2[], int N)
{
    // special case, for only element which is
    // missing in second array
    if (N == 1)
        return arr1[0];
 
    // special case, for first element missing
    if (arr1[0] != arr2[0])
        return arr1[0];
 
    // Initialize current corner points
    int lo = 0,  hi = N - 1;
 
    // loop until lo < hi
    while (lo < hi)
    {
        int mid = (lo + hi) / 2;
 
        // If element at mid indices are equal
        // then go to right subarray
        if (arr1[mid] == arr2[mid])
            lo = mid;
        else
            hi = mid;
 
        // if lo, hi becomes contiguous,  break
        if (lo == hi - 1)
            break;
    }
 
    // missing element will be at hi index of
    // bigger array
    return arr1[hi];
}
 
// This function mainly does basic error checking
// and calls findMissingUtil
void findMissing(int arr1[], int arr2[], int M, int N)
{
    if (N == M-1)
        cout << "Missing Element is "
        << findMissingUtil(arr1, arr2, M) << endl;
    else if (M == N-1)
        cout << "Missing Element is "
        << findMissingUtil(arr2, arr1, N) << endl;
    else
        cout << "Invalid Input";
}
 
// Driver Code
int main()
{
    int arr1[] = {1, 4, 5, 7, 9};
    int arr2[] = {4, 5, 7, 9};
 
    int M = sizeof(arr1) / sizeof(int);
    int N = sizeof(arr2) / sizeof(int);
 
    findMissing(arr1, arr2, M, N);
 
    return 0;
}

Java

// Java program to find missing element
// from same arrays
// (except one missing element)
 
import java.io.*;
class MissingNumber {
 
    /* Function to find missing element based
     on binary search approach. arr1[] is of
     larger size and N is size of it.arr1[] and
     arr2[] are assumed to be in same order. */
    int findMissingUtil(int arr1[], int arr2[],
                        int N)
    {
        // special case, for only element
        // which is missing in second array
        if (N == 1)
            return arr1[0];
 
        // special case, for first
        // element missing
        if (arr1[0] != arr2[0])
            return arr1[0];
 
        // Initialize current corner points
        int lo = 0, hi = N - 1;
 
        // loop until lo < hi
        while (lo < hi) {
            int mid = (lo + hi) / 2;
 
            // If element at mid indices are
            // equal then go to right subarray
            if (arr1[mid] == arr2[mid])
                lo = mid;
            else
                hi = mid;
 
            // if lo, hi becomes
            // contiguous, break
            if (lo == hi - 1)
                break;
        }
 
        // missing element will be at hi
        // index of bigger array
        return arr1[hi];
    }
 
    // This function mainly does basic error
    // checking and calls findMissingUtil
    void findMissing(int arr1[], int arr2[],
                              int M, int N)
    {
        if (N == M - 1)
        System.out.println("Missing Element is "
        + findMissingUtil(arr1, arr2, M) + "\n");
        else if (M == N - 1)
        System.out.println("Missing Element is "
        + findMissingUtil(arr2, arr1, N) + "\n");
        else
        System.out.println("Invalid Input");
    }
 
    // Driver Code
    public static void main(String args[])
    {
        MissingNumber obj = new MissingNumber();
        int arr1[] = { 1, 4, 5, 7, 9 };
        int arr2[] = { 4, 5, 7, 9 };
        int M = arr1.length;
        int N = arr2.length;
        obj.findMissing(arr1, arr2, M, N);
    }
}
 
// This code is contributed by Anshika Goyal.

Python3

# Python3 program to find missing
# element from same arrays
# (except one missing element)
 
# Function to find missing element based
# on binary search approach. arr1[] is
# of larger size and N is size of it.
# arr1[] and arr2[] are assumed
# to be in same order.
def findMissingUtil(arr1, arr2, N):
 
    # special case, for only element
    # which is missing in second array
    if N == 1:
        return arr1[0];
 
    # special case, for first
    # element missing
    if arr1[0] != arr2[0]:
        return arr1[0]
 
    # Initialize current corner points
    lo = 0
    hi = N - 1
     
    # loop until lo < hi
    while (lo < hi):
     
        mid = (lo + hi) / 2
 
        # If element at mid indices
        # are equal then go to
        # right subarray
        if arr1[mid] == arr2[mid]:
            lo = mid
        else:
            hi = mid
 
        # if lo, hi becomes
        # contiguous, break
        if lo == hi - 1:
            break
     
    # missing element will be at
    # hi index of bigger array
    return arr1[hi]
 
# This function mainly does basic
# error checking and calls
# findMissingUtil
def findMissing(arr1, arr2, M, N):
 
    if N == M-1:
        print("Missing Element is",
            findMissingUtil(arr1, arr2, M))
    elif M == N-1:
        print("Missing Element is",
            findMissingUtil(arr2, arr1, N))
    else:
        print("Invalid Input")
 
# Driver Code
arr1 = [1, 4, 5, 7, 9]
arr2 = [4, 5, 7, 9]
M = len(arr1)
N = len(arr2)
findMissing(arr1, arr2, M, N)
 
# This code is contributed by Smitha Dinesh Semwal

C#

// C# program to find missing element from
// same arrays (except one missing element)
using System;
 
class GFG {
 
    /* Function to find missing element based
    on binary search approach. arr1[] is of
    larger size and N is size of it.arr1[] and
    arr2[] are assumed to be in same order. */
    static int findMissingUtil(int []arr1,
                            int []arr2, int N)
    {
         
        // special case, for only element
        // which is missing in second array
        if (N == 1)
            return arr1[0];
 
        // special case, for first
        // element missing
        if (arr1[0] != arr2[0])
            return arr1[0];
 
        // Initialize current corner points
        int lo = 0, hi = N - 1;
 
        // loop until lo < hi
        while (lo < hi) {
            int mid = (lo + hi) / 2;
 
            // If element at mid indices are
            // equal then go to right subarray
            if (arr1[mid] == arr2[mid])
                lo = mid;
            else
                hi = mid;
 
            // if lo, hi becomes
            // contiguous, break
            if (lo == hi - 1)
                break;
        }
 
        // missing element will be at hi
        // index of bigger array
        return arr1[hi];
    }
 
    // This function mainly does basic error
    // checking and calls findMissingUtil
    static void findMissing(int []arr1, int []arr2,
                            int M, int N)
    {
        if (N == M - 1)
            Console.WriteLine("Missing Element is "
            + findMissingUtil(arr1, arr2, M) + "\n");
        else if (M == N - 1)
            Console.WriteLine("Missing Element is "
            + findMissingUtil(arr2, arr1, N) + "\n");
        else
            Console.WriteLine("Invalid Input");
    }
 
    // Driver Code
    public static void Main()
    {
        int []arr1 = { 1, 4, 5, 7, 9 };
        int []arr2 = { 4, 5, 7, 9 };
        int M = arr1.Length;
        int N = arr2.Length;
        findMissing(arr1, arr2, M, N);
    }
}
 
// This code is contributed by anuj_67.

Javascript

    // Javascript program to find missing element from
    // same arrays (except one missing element)
     
    /* Function to find missing element based
    on binary search approach. arr1[] is of
    larger size and N is size of it.arr1[] and
    arr2[] are assumed to be in same order. */
    function findMissingUtil(arr1, arr2, N)
    {
          
        // special case, for only element
        // which is missing in second array
        if (N == 1)
            return arr1[0];
  
        // special case, for first
        // element missing
        if (arr1[0] != arr2[0])
            return arr1[0];
  
        // Initialize current corner points
        let lo = 0, hi = N - 1;
  
        // loop until lo < hi
        while (lo < hi) {
            let mid = parseInt((lo + hi) / 2, 10);
  
            // If element at mid indices are
            // equal then go to right subarray
            if (arr1[mid] == arr2[mid])
                lo = mid;
            else
                hi = mid;
  
            // if lo, hi becomes
            // contiguous, break
            if (lo == hi - 1)
                break;
        }
  
        // missing element will be at hi
        // index of bigger array
        return arr1[hi];
    }
  
    // This function mainly does basic error
    // checking and calls findMissingUtil
    function findMissing(arr1, arr2, M, N)
    {
        if (N == M - 1)
            document.write("Missing Element is "
            + findMissingUtil(arr1, arr2, M) + "</br>");
        else if (M == N - 1)
            document.write("Missing Element is "
            + findMissingUtil(arr2, arr1, N) + "</br>");
        else
            document.write("Invalid Input" + "</br>");
    }
     
    let arr1 = [ 1, 4, 5, 7, 9 ];
    let arr2 = [ 4, 5, 7, 9 ];
    let M = arr1.length;
    let N = arr2.length;
    findMissing(arr1, arr2, M, N);
     

---

输出 ：

Missing Element is 1

如果输入数组的顺序不同怎么办？
在这种情况下，缺少的元素只是两个数组的所有元素的异或。感谢 Yolo Song 提出这个建议。

CPP

// C++ program to find missing element from one array
// such that it has all elements of other array except
// one.  Elements in two arrays can be in any order.
#include 
using namespace std;
 
// This function mainly does XOR of all elements
// of arr1[] and arr2[]
void findMissing(int arr1[], int arr2[], int M,
                 int N)
{
    if (M != N-1 && N != M-1)
    {
        cout << "Invalid Input";
        return;
    }
 
    // Do XOR of all element
    int res = 0;
    for (int i=0; i

Java

// Java program to find missing element
// from one array such that it has all
// elements of other array except one.
// Elements in two arrays can be in any order.
 
import java.io.*;
class Missing {
     
    // This function mainly does XOR of
    // all elements of arr1[] and arr2[]
    void findMissing(int arr1[], int arr2[],
                              int M, int N)
    {
        if (M != N - 1 && N != M - 1) {
        System.out.println("Invalid Input");
            return;
        }
 
        // Do XOR of all element
        int res = 0;
        for (int i = 0; i < M; i++)
            res = res ^ arr1[i];
        for (int i = 0; i < N; i++)
            res = res ^ arr2[i];
 
        System.out.println("Missing element is "
                                          + res);
    }
 
    // Driver Code
    public static void main(String args[])
    {
        Missing obj = new Missing();
        int arr1[] = { 4, 1, 5, 9, 7 };
        int arr2[] = { 7, 5, 9, 4 };
        int M = arr1.length;
        int N = arr2.length;
        obj.findMissing(arr1, arr2, M, N);
    }
}
 
// This code is contributed by Anshika Goyal.

蟒蛇3

# Python 3 program to find
# missing element from one array
# such that it has all elements
# of other array except
# one. Elements in two arrays
# can be in any order.
 
# This function mainly does XOR of all elements
# of arr1[] and arr2[]
def findMissing(arr1,arr2, M, N):
    if (M != N-1 and N != M-1):
     
        print("Invalid Input")
        return
     
 
    # Do XOR of all element
    res = 0
    for i in range(0,M):
        res = res^arr1[i];
    for i in range(0,N):
        res = res^arr2[i]
     
    print("Missing element is",res)
 
# Driver Code
arr1 = [4, 1, 5, 9, 7]
arr2 = [7, 5, 9, 4]
M = len(arr1)
N = len(arr2)
findMissing(arr1, arr2, M, N)
 
# This code is contributed
# by Smitha Dinesh Semwal

C＃

// C# program to find missing element
// from one array such that it has all
// elements of other array except one.
// Elements in two arrays can be in
// any order.
using System;
class GFG {
     
    // This function mainly does XOR of
    // all elements of arr1[] and arr2[]
    static void findMissing(int []arr1,
                            int []arr2,
                            int M, int N)
    {
        if (M != N - 1 && N != M - 1)
        {
            Console.WriteLine("Invalid Input");
            return;
        }
 
        // Do XOR of all element
        int res = 0;
        for (int i = 0; i < M; i++)
            res = res ^ arr1[i];
        for (int i = 0; i < N; i++)
            res = res ^ arr2[i];
 
        Console.WriteLine("Missing element is "
                                        + res);
    }
 
    // Driver Code
    public static void Main()
    {
     
        int []arr1 = {4, 1, 5, 9, 7};
        int []arr2 = {7, 5, 9, 4};
        int M = arr1.Length;
        int N = arr2.Length;
        findMissing(arr1, arr2, M, N);
    }
}
 
// This code is contributed by anuj_67.

Javascript

 
    // Javascript program to find
    // missing element from one array
    // such that it has all elements
    // of other array except one.
    // Elements in two arrays
    // can be in any order.
     
    // This function mainly does XOR
    // of all elements
    // of arr1[] and arr2[]
    function findMissing(arr1, arr2, M, N)
    {
        if (M != N-1 && N != M-1)
        {
            document.write("Invalid Input");
            return;
        }
 
        // Do XOR of all element
        let res = 0;
        for (let i=0; i<M; i++)
           res = res^arr1[i];
        for (let i=0; i<N; i++)
           res = res^arr2[i];
 
        document.write("Missing element is " + res);
    }
     
    let arr1 = [4, 1, 5, 9, 7];
    let arr2 = [7, 5, 9, 4];
   
    let M = arr1.length;
    let N = arr2.length;
   
    findMissing(arr1, arr2, M, N);
     

输出 ：

Missing Element is 1

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