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
.
题解:
题目的要求大概就是判断所要搜索的target是否在所给的矩阵中。要注意的是矩阵每一行从左到右,数字是递增的;每一列从上到下,数字也是递增的。
根据分治算法的思想,一开始可以从矩阵的右上角的元素开始判断,如果target大于该右上角的元素,则在这一行中必定没有target元素;如果target小于该右上角的元素,则在这一列中必定也没有target元素。可以根据这种方法不断缩小查找范围。
具体的code如下:
class Solution {public: bool searchMatrix(vector<vector<int>>& matrix, int target) { if(matrix.size() == 0) return false; int i = 0; int j = matrix[0].size()-1; while(i < matrix.size() && j >= 0) { if(matrix[i][j] == target) return true; if(matrix[i][j] > target) j--; else i++; } return false; }};其实这道题只要知道解题的方法就能轻松地写出代码。
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- 240. Search a 2D Matrix II
- 240. Search a 2D Matrix II
- 240. Search a 2D Matrix II
- 240. Search a 2D Matrix II
- 240. Search a 2D Matrix II
- 240. Search a 2D Matrix II
- 240. Search a 2D Matrix II
- 240. Search a 2D Matrix II
- 240. Search a 2D Matrix II
- 240. Search a 2D Matrix II
- 240. Search a 2D Matrix II
- 240. Search a 2D Matrix II
- 240. Search a 2D Matrix II
- 240. Search a 2D Matrix II
- 240. Search a 2D Matrix II
- 240. Search a 2D Matrix II
- 240. Search a 2D Matrix II
- 240.Search a 2D Matrix II
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- 240. Search a 2D Matrix II
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