LeetCode--240. Search a 2D Matrix II

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.

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

.

Solution：

由题意可知，数组每行、每列都是从小到大有序排列。

从右上角出发，有3种选择：等于target，找到该数，返回true；

小于target，向下移找到更大的数；

大于target，向左移，找到更小的数；

按照上述方法查找，如果超出数组界限，则找不到target，返回false

Code：

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

int i = 0, j = n - 1;
while (i < m && j >= 0) {
if (matrix[i][j] == target){
return true;
} else if (matrix[i][j] > target) {
j--;
} else {
i++;
}
}

return false;
}
};