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
.
如果我们观察题目中给的那个例子,我们可以发现有两个位置的数字很有特点,左下角和右上角的数。左下角的18,往上所有的数变小,往右所有数增加,那么我们就可以和目标数相比较,如果目标数大,就往右搜,如果目标数小,就往左搜。这样就可以判断目标数是否存在。
c++实现:
class Solution {
public:
bool searchMatrix(vector<vector<int>>& matrix, int target) {
if(matrix.size()==0 || matrix[0].size()==0)
return false;
int m=matrix.size();
int n=matrix[0].size();
if (matrix[0][0]>target || matrix[m-1][n-1]<target)
return false;
int x=m-1;
int y=0;
while(x>=0 && y<n){
if (matrix[x][y]<target){ y++; continue;}
if (matrix[x][y]>target){ x--;continue;}
if (matrix[x][y]==target)
return true;
}
return false;
}
};