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【LeetCode】Search a 2D Matrix
Search a 2D Matrix
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.
思路：纵向二分查找确定target所在行，再进行普通的二分查找。
class Solution {public: bool searchMatrix(vector<vector<int> > &matrix, int target) { if(matrix.empty()) return false; else if(matrix[0].empty()) return false; else { int m = matrix.size(); int n = matrix[m-1].size(); //判断越界 if(matrix[0][0] > target || matrix[m-1][n-1] < target) return false; //基于行首的二分查找，定位到行 int low = 0; int high = m-1; int mid; int num; while(low <= high) { mid = (low+high)/2; num = matrix[mid][0]; if(num == target) return true; else if(num > target) high = mid-1; else low = mid+1; } //短路操防止越界 if(low >= m || matrix[low][0] > target) low -= 1; int row = low; //第row行内进行二分查找 low = 0; high = matrix[row].size()-1; while(low <= high) { mid = (low+high)/2; num = matrix[row][mid]; if(num == target) return true; else if(num > target) high = mid-1; else low = mid+1; } return false; } }};
【LeetCode】Search a 2D Matrix
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