LeetCode 240. Search a 2D Matrix II
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题目:
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
.
思路:
1.二分查找第一列找到最高行,二分查找最后一列找到最低行,然后对它们之间每一行进行二分查找 O(m * lgn)
2.直接对每行进行二分查找 O(m * lgn)
3.对于每一个数,上方比它小,右方比它大 O(m * n)
代码 C++ 思路2:
bool bin_search(vector<int>& num , int target){ int start = 0; int end = num.size() - 1; while (start <= end) { int mid = start + (end - start) / 2; if (num[mid] == target) return true; else if (num[mid] > target) end = mid - 1; else if (num[mid] < target) start = mid + 1; } return false; } class Solution { public: bool searchMatrix(vector<vector<int>>& matrix, int target) { for (int i = 0; i <= matrix.size() - 1; i++) { if (bin_search(matrix[i],target)) return true; } return false; } };
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