240. Search a 2D Matrix II

问题描述

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.

题目链接：

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思路分析

给一个二维数组，数组中元素每一行从左到右升序排列；每一列从上倒下升序排列。判断一个数是否在数组中。

从数组的右上角出发，如果比target小就向下移动；如果比target大就向左移动，出现越界就返回false即可。

代码

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

时间复杂度： O(m+n) 空间复杂度： O(1)

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反思

tag是分治，不太明白。