算法：LeetCode240

2019-11-06 07:21:41

字体：大中小

来源：转载

供稿：网友

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.

解题思路

分治法求解，首先根据矩阵的特点，横向和纵向是分别升序的。可以从右上角或者左下角进行判断。拿左下角来说 1. 如果比target大，那么需要往上走，因为当前行的右边都比它大 2. 如果比target小，那么需要往右走，因为当前列的上面比它都小 3. 相等则命中

每次都是通过减少一行或者一列来向前搜索，复杂度就是O（M+N）M,N为矩阵行列数

代码

class Solution {public: bool searchMatrix(vector<vector<int>>& matrix, int target) { int row=matrix.size(); if(row == 0) return false; int column=matrix[0].size(); int j=0; int i=row-1; while(j<column&&i>=0) { int temp=matrix[i][j]; if(temp==target) { return true; } if(temp>target) { i--; } else { j++; } } return false; }};

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