二维数组的搜索算法（矩阵）
二维数组中的线性搜索
线性搜索是一种简单的顺序搜索算法。它用于通过遍历数组中的每个元素来查找数组中是否存在特定元素。虽然在二维数组中搜索是完全相同的，但这里需要遍历所有单元格，这样，在二维数组中搜索任何元素。
下面是二维数组中线性搜索的实现
C++
// C++ code for the above approach
#include
using namespace std;
vector linearSearch(vector> arr, int target)
{
for (int i = 0; i < arr.size(); i++) {
for (int j = 0; j < arr[i].size(); j++) {
if (arr[i][j] == target) {
return {i, j};
}
}
}
return {-1, -1};
}
// Driver code
int main()
{
vector> arr = { { 3, 12, 9 },
{ 5, 2, 89 },
{ 90, 45, 22 } };
int target = 89;
vector ans = linearSearch(arr, target);
cout << "Element found at index: [" << ans[0] << " " < Java
// Linear Search in 2D arrays
import java.util.Arrays;
public class GFG {
public static void main(String[] args)
{
int arr[][] = { { 3, 12, 9 },
{ 5, 2, 89 },
{ 90, 45, 22 } };
int target = 89;
int ans[] = linearSearch(arr, target);
System.out.println("Element found at index: "
+ Arrays.toString(ans));
}
static int[] linearSearch(int[][] arr, int target)
{
for (int i = 0; i < arr.length; i++) {
for (int j = 0; j < arr[i].length; j++) {
if (arr[i][j] == target) {
return new int[] { i, j };
}
}
}
return new int[] { -1, -1 };
}
}Python3
# Python code for the above approach
def linearSearch (arr, target):
for i in range(len(arr)):
for j in range(len(arr[i])):
if (arr[i][j] == target):
return [i, j]
return [-1, -1]
# Driver code
arr = [[3, 12, 9], [5, 2, 89], [90, 45, 22]]
target = 89
ans = linearSearch(arr, target)
print(f"Element found at index: [{ans[0]} {ans[1]}]")
# This code is contributed by Saurabh JaiswalC#
// Linear Search in 2D arrays
using System;
public class GFG {
public static void Main(string[] args)
{
int[, ] arr = { { 3, 12, 9 },
{ 5, 2, 89 },
{ 90, 45, 22 } };
int target = 89;
int[] ans = linearSearch(arr, target);
Console.WriteLine("Element found at index: ["
+ ans[0] + "," + ans[1]+"]");
}
static int[] linearSearch(int[, ] arr, int target)
{
for (int i = 0; i < arr.GetLength(0); i++) {
for (int j = 0; j < arr.GetLength(1); j++) {
if (arr[i, j] == target) {
return new int[] { i, j };
}
}
}
return new int[] { -1, -1 };
}
}
// This code is contributed by ukasp.Javascript
C++
// Binary Search on sorted 2D array
#include
using namespace std;
vector findAns(vector> arr, int target)
{
int r = 0;
int c = arr[r].size() - 1;
while (r < arr.size() && c >= 0) {
if (arr[r] == target) {
return { r, c };
}
// Target lies in further row
if (arr[r] < target) {
r++;
}
// Target lies in previous column
else {
c--;
}
}
return { -1, -1 };
}
// Driver Code
int main()
{
// Binary search in sorted matrix
vector> arr = { { 1, 2, 3, 4 },
{ 5, 6, 7, 8 },
{ 9, 10, 11, 12 } };
vector ans = findAns(arr, 12);
cout << "Element found at index: [";
for(int i = 0; i < ans.size(); i++){
if(i == ans.size() - 1)
cout << ans[i];
else
cout << ans[i] << ", ";
}
cout << "]";
}
// This code is contributed by Samim Hossain Mondal. Java
// Binary Search on sorted 2D array
import java.util.Arrays;
class GFG {
static int[] findAns(int[][] arr, int target)
{
int r = 0;
int c = arr[r].length - 1;
while (r < arr.length && c >= 0) {
if (arr[r] == target) {
return new int[] { r, c };
}
// Target lies in further row
if (arr[r] < target) {
r++;
}
// Target lies in previous column
else {
c--;
}
}
return new int[] { -1, -1 };
}
// Driver Code
public static void main(String[] args)
{
// Binary search in sorted matrix
int arr[][] = { { 1, 2, 3, 4 },
{ 5, 6, 7, 8 },
{ 9, 10, 11, 12 } };
int[] ans = findAns(arr, 12);
System.out.println("Element found at index: "
+ Arrays.toString(ans));
}
}Python3
# Binary Search on sorted 2D array
def findAns(arr, target):
r = 0;
c = len(arr[r]) - 1;
while (r < len(arr) and c >= 0):
if (arr[r] == target):
return [r, c];
# Target lies in further row
if (arr[r] < target):
r += 1;
# Target lies in previous column
else:
c -=1;
return [ -1, -1];
# Driver Code
if __name__ == '__main__':
# Binary search in sorted matrix
arr = [[1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12]];
ans = findAns(arr, 12);
print("Element found at index: ", ans);
# This code contributed by shikhasingrajputC#
// Binary Search on sorted 2D array
using System;
class GFG {
static int[] findAns(int[, ] arr, int target)
{
int r = 0;
int c = arr.GetLength(1) - 1;
while (r < arr.GetLength(0) && c >= 0) {
if (arr[r, c] == target) {
return new int[] { r, c };
}
// Target lies in further row
if (arr[r, c] < target) {
r++;
}
// Target lies in previous column
else {
c--;
}
}
return new int[] { -1, -1 };
}
// Driver Code
public static void Main(string[] args)
{
// Binary search in sorted matrix
int[, ] arr = { { 1, 2, 3, 4 },
{ 5, 6, 7, 8 },
{ 9, 10, 11, 12 } };
int[] ans = findAns(arr, 12);
Console.Write("Element found at index: [");
int i = 0;
for (i = 0; i < ans.Length - 1; i++)
Console.Write(ans[i] + " ,");
Console.Write(ans[i] + "]");
}
}
// This code is contributed by ukasp.Javascript
输出
Element found at index: [1, 2]二维数组中线性搜索的时间复杂度为O (N * M) ，其中N是行数， M是列数。
二进制搜索 二维阵列
二进制搜索是在数组中搜索的一种有效方法。二进制搜索适用于排序数组。在每次迭代中，搜索空间被分成两半，这就是为什么二分搜索比线性搜索更有效的原因。
为什么二进制搜索对于在未排序的数组中搜索没有用？
在任何算法中应用二分搜索的基本条件是搜索空间应该是有序的。要在二维数组中执行二分搜索，需要对数组进行排序。这里给出了一个未排序的二维数组，因此无法在未排序的数组中应用二分搜索。要首先应用二分搜索，二维数组需要按照本身需要(M*N)log(M*N)时间的任何顺序进行排序。所以在这里搜索任何元素的总时间复杂度是O((M * N) log(M * N)) + O(N + M)与线性搜索的时间复杂度相比非常差，只有O (N*M) 。因此，线性搜索用于在未排序的数组中进行搜索，而不是二分搜索。
下面是二维数组中二分查找的实现
C++
// Binary Search on sorted 2D array
#include
using namespace std;
vector findAns(vector> arr, int target)
{
int r = 0;
int c = arr[r].size() - 1;
while (r < arr.size() && c >= 0) {
if (arr[r] == target) {
return { r, c };
}
// Target lies in further row
if (arr[r] < target) {
r++;
}
// Target lies in previous column
else {
c--;
}
}
return { -1, -1 };
}
// Driver Code
int main()
{
// Binary search in sorted matrix
vector> arr = { { 1, 2, 3, 4 },
{ 5, 6, 7, 8 },
{ 9, 10, 11, 12 } };
vector ans = findAns(arr, 12);
cout << "Element found at index: [";
for(int i = 0; i < ans.size(); i++){
if(i == ans.size() - 1)
cout << ans[i];
else
cout << ans[i] << ", ";
}
cout << "]";
}
// This code is contributed by Samim Hossain Mondal.
Java
// Binary Search on sorted 2D array
import java.util.Arrays;
class GFG {
static int[] findAns(int[][] arr, int target)
{
int r = 0;
int c = arr[r].length - 1;
while (r < arr.length && c >= 0) {
if (arr[r] == target) {
return new int[] { r, c };
}
// Target lies in further row
if (arr[r] < target) {
r++;
}
// Target lies in previous column
else {
c--;
}
}
return new int[] { -1, -1 };
}
// Driver Code
public static void main(String[] args)
{
// Binary search in sorted matrix
int arr[][] = { { 1, 2, 3, 4 },
{ 5, 6, 7, 8 },
{ 9, 10, 11, 12 } };
int[] ans = findAns(arr, 12);
System.out.println("Element found at index: "
+ Arrays.toString(ans));
}
}
Python3
# Binary Search on sorted 2D array
def findAns(arr, target):
r = 0;
c = len(arr[r]) - 1;
while (r < len(arr) and c >= 0):
if (arr[r] == target):
return [r, c];
# Target lies in further row
if (arr[r] < target):
r += 1;
# Target lies in previous column
else:
c -=1;
return [ -1, -1];
# Driver Code
if __name__ == '__main__':
# Binary search in sorted matrix
arr = [[1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12]];
ans = findAns(arr, 12);
print("Element found at index: ", ans);
# This code contributed by shikhasingrajput
C＃
// Binary Search on sorted 2D array
using System;
class GFG {
static int[] findAns(int[, ] arr, int target)
{
int r = 0;
int c = arr.GetLength(1) - 1;
while (r < arr.GetLength(0) && c >= 0) {
if (arr[r, c] == target) {
return new int[] { r, c };
}
// Target lies in further row
if (arr[r, c] < target) {
r++;
}
// Target lies in previous column
else {
c--;
}
}
return new int[] { -1, -1 };
}
// Driver Code
public static void Main(string[] args)
{
// Binary search in sorted matrix
int[, ] arr = { { 1, 2, 3, 4 },
{ 5, 6, 7, 8 },
{ 9, 10, 11, 12 } };
int[] ans = findAns(arr, 12);
Console.Write("Element found at index: [");
int i = 0;
for (i = 0; i < ans.Length - 1; i++)
Console.Write(ans[i] + " ,");
Console.Write(ans[i] + "]");
}
}
// This code is contributed by ukasp.
Javascript
输出
Element found at index: [2, 3]在已排序的二维数组中，二分查找的时间复杂度为O (N + M) ，其中N是行数， M是列数。