LeetCode 240. Search a 2D Matrix II

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 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.

分析：

可从二维数组最后一行第一列的数开始与target比较，当matrix[m][n] > target时，m减1，表示该行的数都比target大，不用继续比较该行的数；当matrix[m][n] < target时，n加1，表示该列的数都比target小，不用继续比较该列的数，应比较更大的列；当找到相等的数时，返回true；否则循环结束后返回false。

代码如下：

class Solution {
public:
	bool searchMatrix(vector<vector<int>>& matrix, int target) {
		int m = matrix.size() - 1;
		int n = 0;
		while (m >= 0 && n < matrix[m].size()) {
			if (matrix[m][n] == target) return true;
			if (matrix[m][n] > target) m--;
			else n++;
		}
		return false;
	}
};