Leetcode-456. 132 Pattern

前言：为了后续的实习面试，开始疯狂刷题，非常欢迎志同道合的朋友一起交流。因为时间比较紧张，目前的规划是先过一遍，写出能想到的最优算法，第二遍再考虑最优或者较优的方法。如有错误欢迎指正。博主首发CSDN，mcf171专栏。

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Given a sequence of n integers a1, a2, ..., an, a 132 pattern is a subsequence ai, aj, ak such that i < j < k and ai < ak < aj. Design an algorithm that takes a list of n numbers as input and checks whether there is a 132 pattern in the list.

Note: n will be less than 15,000.

Example 1:

Input: [1, 2, 3, 4]Output: FalseExplanation: There is no 132 pattern in the sequence.

Example 2:

Input: [3, 1, 4, 2]Output: TrueExplanation: There is a 132 pattern in the sequence: [1, 4, 2].

Example 3:

Input: [-1, 3, 2, 0]Output: TrueExplanation: There are three 132 patterns in the sequence: [-1, 3, 2], [-1, 3, 0] and [-1, 2, 0].

这个题目可以用暴力的方法解，优化的办法暂时没想到。时间复杂度O(n^2)Your runtime beats 0.08% of java submissions.

public class Solution {    public boolean find132pattern(int[] nums) {        if(nums.length < 3) return false;        boolean flag = false;        for(int k = nums.length - 1; k >=2; k --){            int i = 0, j = k -1;            while(i < j){                if(nums[i] < nums[k] && nums[k] < nums[j]){flag = true;break;}                if(nums[i] >=nums[k]) i ++;                if(nums[j] <= nums[k]) j--;            }            if(flag) break;        }        return flag;    }}