README_EN.md

74. Search a 2D Matrix

中文文档

Description

Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties:

• Integers in each row are sorted from left to right.
• The first integer of each row is greater than the last integer of the previous row.

 

Example 1:

Input: matrix = [[1,3,5,7],[10,11,16,20],[23,30,34,60]], target = 3
Output: true

Example 2:

Input: matrix = [[1,3,5,7],[10,11,16,20],[23,30,34,60]], target = 13
Output: false

 

Constraints:

• m == matrix.length
• n == matrix[i].length
• 1 <= m, n <= 100
• -104 <= matrix[i][j], target <= 104

Solutions

Python3

class Solution:
    def searchMatrix(self, matrix: List[List[int]], target: int) -> bool:
        m, n = len(matrix), len(matrix[0])
        left, right = 0, m * n - 1
        while left < right:
            mid = (left + right) >> 1
            x, y = divmod(mid, n)
            if matrix[x][y] >= target:
                right = mid
            else:
                left = mid + 1
        return matrix[left // n][left % n] == target

Java

class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {
        int m = matrix.length, n = matrix[0].length;
        int left = 0, right = m * n - 1;
        while (left < right) {
            int mid = (left + right) >> 1;
            int x = mid / n, y = mid % n;
            if (matrix[x][y] >= target) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        return matrix[left / n][left % n] == target;
    }
}

C++

class Solution {
public:
    bool searchMatrix(vector<vector<int>>& matrix, int target) {
        int m = matrix.size(), n = matrix[0].size();
        int left = 0, right = m * n - 1;
        while (left < right) {
            int mid = left + right >> 1;
            int x = mid / n, y = mid % n;
            if (matrix[x][y] >= target) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        return matrix[left / n][left % n] == target;
    }
};

JavaScript

/**
 * @param {number[][]} matrix
 * @param {number} target
 * @return {boolean}
 */
var searchMatrix = function(matrix, target) {
    const m = matrix.length;
    const n = matrix[0].length;
    let left = 0;
    let right = m * n - 1;
    while (left < right) {
        const mid = (left + right) >> 1;
        const x = Math.floor(mid / n);
        const y = mid % n;
        if (matrix[x][y] >= target) {
            right = mid;
        } else {
            left = mid + 1;
        }
    }
    return matrix[Math.floor(left / n)][left % n] == target;
};

Go

func searchMatrix(matrix [][]int, target int) bool {
	m, n := len(matrix), len(matrix[0])
	left, right := 0, m*n-1
	for left < right {
		mid := (left + right) >> 1
		x, y := mid/n, mid%n
		if matrix[x][y] >= target {
			right = mid
		} else {
			left = mid + 1
		}
	}
	return matrix[left/n][left%n] == target
}

...