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.
从右上角开始,若目标值比该值大,行++,否则列–
class Solution {
public:
bool searchMatrix(vector<vector<int>>& matrix, int target) {
if(matrix.size() == 0)
return false;
int i = 0,j = matrix[0].size() - 1;
while(i < matrix.size() && j < matrix[0].size()){
if(matrix[i][j] == target)
return true;
if(matrix[i][j] < target)
i++;
else
j--;
}
return false;
}
};
本文介绍了一种高效的矩阵搜索算法,用于在一个特殊排列的二维矩阵中查找特定值。此矩阵的每一行从左到右递增排序,每一列从上到下递增排序。文章通过实例详细解释了算法的工作原理,并提供了一个C++实现示例。
369

被折叠的 条评论
为什么被折叠?



