[LeetCode74]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

.

因为题目所给的矩阵每行已经从左到右排好序，并且每一行的第一个元素都比上一行的最后一个元素大，因此只要将目标值与每一行的首尾元素进行比较就能确定目标值所在的行，然后在这一行上运用二分查找。时间复杂度为O(mlogn)。

class Solution {
public:
bool searchMatrix(vector<vector<int>>& matrix, int target) {
bool r=false;
//vector<int>::iterator it;
for(int i=0;i<matrix.size();i++)
{
if(target>=matrix[i][0]&&target<=matrix[i][matrix[i].size()-1])
{
int low=0,high=matrix[i].size()-1,mid;
while(low<=high)
{
mid=low+(high-low)/2;
if(target==matrix[i][mid])
{
r=true;
break;
}
else if(target>matrix[i][mid])
{
low=mid+1;
}
else
{
high=mid-1;
}
}
}
}
return r;
}
};