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.
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] ]
Example 1:
Input: matrix, target = 5
Output: true
Example 2:
Input: matrix, target = 20 Output: false
思路：
充分利用有序。
列是有序的，那么只有当要查找的数>=一行的第一个数，且<=最后一个数时才可能位于这一行。此时利用行有序进行二分查找。
class Solution {
public:
bool searchMatrix(vector<vector<int>>& matrix, int target) {
if(matrix.empty() || matrix[0].empty())
return false;
int m = matrix.size();
int n = matrix[0].size();
for(int i=0;i<m;i++){
if(matrix[i][0]<=target){
if(matrix[i][n-1]>=target){//如果在这个一行则进行二分查找
int left=0;
int right=n-1;
while(left<=right){
int mid=left+(right-left)/2;
if(target < matrix[i][mid]) right=mid-1;
else if(target < matrix[i][mid]) left=mid+1;
else return true;
}
}
}
}
return false;
}
};