[leetcode] 74. Search a 2D Matrix

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 from left to right.
The first integer of each row is greater than the last integer of the previous row.

For example,

Consider the following matrix:

[
[1,   3,  5,  7],
[10, 11, 16, 20],
[23, 30, 34, 50]
]

Given target = 

3

, return 

true

.

解法一：

还是用二分法来做，用到的就是讲一个数值转化成矩阵元素的下表。另外要注意一些corner case，target小于matrix中第一个值或者大于matrix中最后一个值。

class Solution {
public:
bool searchMatrix(vector<vector<int>>& matrix, int target) {
int rows = matrix.size(), cols = matrix[0].size()?matrix[0].size():0;
if(!rows || !cols) return 0;
if(target<matrix[0][0]) return false;
if(target>matrix.back().back()) return false;
int left = 0, right = rows*cols - 1;

while(left<=right){
int mid = left + (right-left)/2;
int i_mid = mid/cols, j_mid = mid%cols;
if(matrix[i_mid][j_mid]==target) return true;
else if(matrix[i_mid][j_mid]>target) right = mid-1;
else left = mid+1;
}

return false;

}
};