Question

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 in ascending from left to right.
Integers in each column are sorted in ascending from top to bottom.

For example,
Consider the following matrix:
[
[1, 4, 7, 11, 15],
[2, 5, 8, 12, 19],
[3, 6, 9, 16, 22],
[10, 13, 14, 17, 24],
[18, 21, 23, 26, 30]
]
Given target = 5, return true.
Given target = 20, return false.

Code

public class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {
        if (matrix == null || matrix.length == 0 || matrix[0].length == 0) return false;
        
        int i = 0, j = matrix[0].length - 1;
        while (i < matrix.length && j >= 0) {
            if (matrix[i][j] == target) return true;
            else if (matrix[i][j] < target) i++;
            else j--;
        }
        return false;
    }
}

Solution

分治法。从矩阵的右上角开始遍历。
（1）该元素与target相同：返回true；
（2）该元素小于target：说明target不可能在这一行，i++；
（3）该元素大于target：说明target不可能在这一列，j--。