【Leetcode】Search a 2D Matrix
LeetCode search 2D matrix
2023-09-14 09:08:07 时间
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:因为题目的条件是第x+1行的第一个数比第x行最后一个数大,所以能够使用二分查找;假设题目条件仅仅说逐行递增,逐列递增,能够从右上角開始查找。此方法也适用于本题。
代码一:
class Solution { public: bool searchMatrix(vector<vector<int> > &matrix, int target) { const int m = matrix.size(); if(m == 0) return false; const int n = matrix[0].size(); if(n == 0) return false; int first = 0; int last = m * n - 1; while(first <= last) { int middle = (first + last) / 2; int val = matrix[middle / n][middle % n]; if(val == target) return true; else if(val < target) first = middle + 1; else last = middle - 1; } return false; } };
代码二:
class Solution { public: bool searchMatrix(vector<vector<int> > &matrix, int target) { const int m = matrix.size(); if(m == 0) return false; const int n = matrix[0].size(); if(n == 0) return false; int row = 0; int column = n - 1; while(row < m && column >= 0) { if(matrix[row][column] == target) return true; else if(matrix[row][column] > target) column--; else row++; } return false; } };
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