題目
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.
分析
思路就是二分法饶囚，不同的是查找時(shí)需要將元素的序號(hào)映射為其在矩陣中的坐標(biāo)每辟。
實(shí)現(xiàn)
class Solution {
public:
bool searchMatrix(vector<vector<int>>& matrix, int target) {
m=matrix.size();
n=matrix[0].size();
return helper(matrix, target, 0, m*n-1);
}
private:
int m,n;
bool helper(vector<vector<int>>& matrix, int target, int start, int end){
int mid = (start + end) / 2;
int x = mid / n;
int y = mid % n;
if(start>=end) return matrix[x][y]==target;
if(matrix[x][y]<target)
return helper(matrix, target, mid+1, end);
else
return helper(matrix, target, start, mid);
}
};
思考
不適用遞歸會(huì)更快。
