[leetcode]456. 132 Pattern

题目链接:https://leetcode.com/problems/132-pattern/#/solutions

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].

We want to search for a subsequence (s1,s2,s3)

INTUITION: The problem would be simpler if we want to find sequence with s1 > s2 > s3, we just need to find s1, followed by s2 and s3. Now if we want to find a 132 sequence, we need to switch up the order of searching. we want to first find s2, followed by s3, then s1.

IDEA: We can start from either side but I think starting from the end allow us to finish in a single pass. The idea is to start from end and search for a candidate for s2 and s3. A number becomes a candidate for s3 if there is any number on the left of s2 that is bigger than it.

DETECTION: Keep track of the largest candidate of s3 and once we encounter any number smaller than s3, we know we found a valid sequence since s1 < s3 implies s1 < s2.

IMPLEMENTATION:

1. Have a stack, each time we store a new number, we first pop out all numbers that are smaller than that number. The numbers that are popped out becomes candidate for s3.
2. We keep track of the maximum of such s3 (which is always the most recently popped number from the stack).
3. Once we encounter any number smaller than s3, we know we found a valid sequence since s1 < s3 implies s1 < s2.

RUNTIME: Each item is pushed and popped once at most, the time complexity is therefore O(n).

EXAMPLE:
i = 6, nums = [ 9, 11, 8, 9, 10, 7, 9 ], S1 candidate = 9, S3 candidate = None, Stack = Empty
i = 5, nums = [ 9, 11, 8, 9, 10, 7, 9 ], S1 candidate = 7, S3 candidate = None, Stack = [9]
i = 4, nums = [ 9, 11, 8, 9, 10, 7, 9 ], S1 candidate = 10, S3 candidate = None, Stack = [9,7]
i = 3, nums = [ 9, 11, 8, 9, 10, 7, 9 ], S1 candidate = 9, S3 candidate = 9, Stack = [10]
i = 2, nums = [ 9, 11, 8, 9, 10, 7, 9 ], S1 candidate = 8, S3 candidate = 9, Stack = [10,9] We have 8<9, sequence found!

class Solution {
public:
    bool find132pattern(vector<int>& nums)
    {
        int s3=INT32_MIN;
        stack<int> st; //大小递减队列
        for(int i=nums.size()-1;i>=0;i--)
        {
            if(nums[i]<s3) return true;
            else
                while(!st.empty() && nums[i]>st.top())
                {
                    s3=st.top();
                    st.pop();
                }
            st.push(nums[i]);
        }
        return false;
    }
};