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.

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.

 

观察到，左下角的数字，向上递减，向右递增，可以从左下角开始查找，循环成立条件是数组下标不越界。如果 当前数 < target，向右查找；如果 当前数 < target，向上查找。

time: O(M+N), space: O(1)

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