题目链接
https://leetcode.com/problems/search-a-2d-matrix-ii/description/
题目描述
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.
题目分析
这个题目很好理解,就是查找在一个矩阵中有没有给定的目标。并且这个矩阵很特殊,每一行从左到右每一列从上到下都是递增的。因为每一行都是有序的,不难想到对每一行使用二分查找,这样时间复杂度为O(mlogn),这个时间复杂度可以接受。不过,由于给定矩阵十分特殊,我们可以考虑从左下角(右上角也有类似结果,这里就不赘述)的元素开始考虑,利用分治算法可以把时间复杂度降到O(m+n).
方法一:对每一行二分查找
class Solution {
public:
bool searchMatrix(vector<vector<int>>& matrix, int target) {
int m = matrix.size();
if (m == 0) return false;
int n = matrix[0].size();
if (n == 0) return false;
for (int i = 0; i < m; i++) {
int left = 0, right = n - 1;
while (left <= right) {
int middle = (left + right) / 2;
if (matrix[i][middle] == target)
return true;
else if (matrix[i][middle] < target)
left = middle + 1;
else
right = middle - 1;
}
if (left < n && matrix[i][left] == target)
return true;
}
return false;
}
};
方法二:分治算法
算法描述
左下角的元素是这一行中最小的元素,同时又是这一列中最大的元素。比较左下角元素和目标:
若左下角元素等于目标,则找到
若左下角元素大于目标,则目标不可能存在于当前矩阵的最后一行,问题规模可以减小为在去掉最后一行的子矩阵中寻找目标
若左下角元素小于目标,则目标不可能存在于当前矩阵的第一列,问题规模可以减小为在去掉第一列的子矩阵中寻找目标
若最后矩阵减小为空,则说明不存在
class Solution {
public:
bool searchMatrix(vector<vector<int>>& matrix, int target) {
int m = matrix.size();
if (m == 0) return false;
int n = matrix[0].size();
if (n == 0) return false;
int i = m - 1, j = 0;
while (i >= 0 && j < n) {
if (matrix[i][j] == target)
return true;
else if (matrix[i][j] < target)
j++;
else
i--;
}
return false;
}
};