[240]Search a 2D Matrix II

【题目描述】

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】

如果从第一个元素开始实现比较麻烦，转换思路，从右上角的元素开始，发现不管哪个元素其左边的元素都比它小，下边的元素都比它大，如果target大于该元素，则搜索其下边的元素，如果小于则搜索其左边的元素。

【代码】

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