Question
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.
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
.
Difficulty:Medium
Category:
Solution
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,就往上搜。这样就可以判断目标数是否存在。当然我们也可以把起始数放在右上角,往左和下搜,停止条件设置正确就行.
class Solution {
public:
bool searchMatrix(vector<vector<int>>& matrix, int target) {
if (matrix.empty() || matrix[0].empty()) return false;
if (target < matrix[0][0] || target > matrix.back().back()) return false;
int row = matrix.size() - 1, col = 0;
while (col < matrix[0].size() && row >= 0) {
int val = matrix[row][col];
if (val == target) return true;
val > target ? --row : ++col;
}
return false;
}
};