132 Pattern

版权声明：本文为博主原创文章，未经博主允许不得转载。 https://blog.csdn.net/zjucor/article/details/82147795

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: False

Explanation: There is no 132 pattern in the sequence.

Example 2:

Input: [3, 1, 4, 2]

Output: True

Explanation: There is a 132 pattern in the sequence: [1, 4, 2].

Example 3:

Input: [-1, 3, 2, 0]

Output: True

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

如果是123pattern，左右2边分别找最大最小就好了，但是这里是132pattern，就是23部分反过来了，比较容易想到用stack也反过来，但是貌似思路还不是很明显

正解：把s2,s3放在一起考虑，每次出现s2>s3的pair就更新s3可能的最大值，这样只需要出现s1<s3就可以判断存在132pattern了。那剩下的问题就是怎么更新s3可能的最大值呢？把s3可能达到的最大值放到stack里，没来一个s2就从stack中pop出比s2小的数，求pop过程最大的数即可

import java.util.TreeMap;

class Solution {
    public boolean find132pattern(int[] nums) {
    	int n = nums.length;
    	if(n < 3)	return false;
    	int min = nums[0];
    	TreeMap<Integer, Integer> s = new TreeMap<Integer, Integer>();
    	for(int i=2; i<n; i++) s.put(nums[i], s.containsKey(nums[i])?1+s.get(nums[i]):1);
    	
    	for(int i=1; i<n-1; i++) {
    		if(nums[i] > min) {
    			Integer t = s.ceilingKey(min+1);
    			if(t != null && t < nums[i])	return true;
    		} else {
    			min = nums[i];
    		}
    		
    		s.put(nums[i+1], s.get(nums[i+1])-1);
    		if(s.get(nums[i+1]) == 0)		s.remove(nums[i+1]);
    	}
    	
    	return false;
    }
}

class Solution:
    def find132pattern(self, nums):
        """
        :type nums: List[int]
        :rtype: bool
        """
        if len(nums)<3: return False
        s3_max=float('-inf')
        st=[]
        for i in nums[::-1]:
            if s3_max>i: return True
            while st and i>st[-1]:
                s3_max=max(s3_max, st.pop())
            st.append(i)
        return False

https://leetcode.com/problems/132-pattern/discuss/94071/Single-pass-C++-O(n)-space-and-time-solution-(8-lines)-with-detailed-explanation.