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
.
在一个给定的矩阵里查找某一个值,矩阵有如下特点:每一行里的整数都是按从左到右排序;每一行的第一数都大于前一行的最后一个数。
解法:二分法Binary Search,利用给定矩阵的特点,Z形大小顺序排列,把矩阵变成数组,使用二分法查找,或者直接在矩阵上使用二分法。
Python:
class Solution(object): def searchMatrix(self, matrix, target): """ :type matrix: List[List[int]] :type target: int :rtype: bool """ if not matrix: return False m, n = len(matrix), len(matrix[0]) left, right = 0, m * n while left < right: mid = left + (right - left) / 2 if matrix[mid / n][mid % n] >= target: right = mid else: left = mid + 1 return left < m * n and matrix[left / n][left % n] == target
C++: Time: O(logm + logn), Space: O(1)
class Solution { public: bool searchMatrix(vector<vector<int>>& matrix, int target) { if (matrix.empty()) { return false; } // Treat matrix as 1D array. const int m = matrix.size(); const int n = matrix[0].size(); int left = 0; int right = m * n - 1; // Find min of left s.t. matrix[left / n][left % n] >= target while (left <= right) { int mid = left + (right - left) / 2; if (matrix[mid / n][mid % n] >= target) { right = mid - 1; } else { left = mid + 1; } } // Check if matrix[left / n][left % n] equals to target. if (left != m * n && matrix[left / n][left % n] == target) { return true; } return false; } };
C++:
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 m = matrix.size(), n = matrix[0].size(); int left = 0, right = m * n - 1; while (left <= right) { int mid = (left + right) / 2; if (matrix[mid / n][mid % n] == target) return true; else if (matrix[mid / n][mid % n] < target) left = mid + 1; else right = mid - 1; } return false; } };