74 Search a 2D Matrix 搜索二维矩阵
Description:
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.
Example:
Example 1:
Input:
matrix = [
[1, 3, 5, 7],
[10, 11, 16, 20],
[23, 30, 34, 50]
]
target = 3
Output: true
Example 2:
Input:
matrix = [
[1, 3, 5, 7],
[10, 11, 16, 20],
[23, 30, 34, 50]
]
target = 13
Output: false
题目描述:
编写一个高效的算法来判断 m x n 矩阵中，是否存在一个目标值。该矩阵具有如下特性：
每行中的整数从左到右按升序排列。
每行的第一个整数大于前一行的最后一个整数。
示例 :
示例 1:
输入:
matrix = [
[1, 3, 5, 7],
[10, 11, 16, 20],
[23, 30, 34, 50]
]
target = 3
输出: true
示例 2:
输入:
matrix = [
[1, 3, 5, 7],
[10, 11, 16, 20],
[23, 30, 34, 50]
]
target = 13
输出: false
思路:
可以将二维矩阵看作一行数组
元素用 [index / matrix[0].size()][index % matrix[0].size()]定位
用二分查找即可
时间复杂度O(lg(mn)), 空间复杂度O(1)
代码:
C++:
class Solution
{
public:
bool searchMatrix(vector>& matrix, int target)
{
if (matrix.empty() or matrix[0].empty()) return false;
int left = 0, right = matrix.size() * matrix[0].size() - 1, n = matrix[0].size();
while (left <= right)
{
int mid = left + ((right - left) >> 1);
int cur = matrix[mid / n][mid % n];
if (target == cur) return true;
else if (target > cur) left = mid + 1;
else right = mid - 1;
}
return false;
}
};
Java:
class Solution {
public boolean searchMatrix(int[][] matrix, int target) {
if (matrix.length == 0 || matrix[0].length == 0) return false;
int left = 0, right = matrix.length * matrix[0].length - 1, n = matrix[0].length;
while (left <= right) {
int mid = left + ((right - left) >> 1);
int cur = matrix[mid / n][mid % n];
if (target == cur) return true;
else if (target > cur) left = mid + 1;
else right = mid - 1;
}
return false;
}
}
Python:
class Solution:
def searchMatrix(self, matrix: List[List[int]], target: int) -> bool:
return target in set(chain(*matrix))