LeetCode 240 Search a 2D Matrix II

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

.
[思路]

从右上角开始, 比较target 和 matrix[m]
的值. 如果matrix[m]
小于target, 则该行不可能有此数,  所以m++，如果matrix[m]
大于target，则该列不可能有此数,
所以n--，遇到边界则表明该矩阵不含target。感谢西施豆腐渣。查找从右上角开始处理，是个很妙的思路。时间复杂度为O(m+n)。

public boolean searchMatrix(int[][] matrix, int target) {
if (matrix.length == 0) return false;
int m = 0, n = matrix[0].length - 1;
while (m < matrix.length && n >= 0) {
int x = matrix[m]
;
if (target == x) return true;
else if (target < x) n--;
else m++;
}
return false;
}