Leetcode 240. Search a 2D Matrix II

原题：
 
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

.
 
解决方法：
由题意可知，每一行或每一列是有序的，可以采用十字交叉搜索的方法，从右上角的点开始：
- 如果相等，则可以直接返回；
- 如果该点的值小于目标值，则表明在前面的列，j--。
-如果该点的值大于目标值，则表明在后面的行，i++。

 
代码：
bool searchMatrix(vector<vector<int>>& matrix, int target) {
int row = matrix.size();
if (row < 1)
return false;
int col = matrix[0].size();
if (col < 1)
return false;

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