456. 132 Pattern

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].

https://discuss.leetcode.com/topic/67881/single-pass-c-o-n-space-and-time-solution-8-lines-with-detailed-explanation

Solution：Stack

思路：
Time Complexity: O(N) Space Complexity: O(N)

Solution Code：

// 定义顺序：s1, s2, s3
// 大小顺序：s1 < s3 < s2
class Solution {
    public boolean find132pattern(int[] nums) {
        int s3 = Integer.MIN_VALUE;
        Deque stack = new ArrayDeque<>();
        for(int i = nums.length - 1; i >= 0; i--) {
            if(nums[i] < s3) return true;
            else {
                // 单调递减栈
                while(!stack.isEmpty() && nums[i] > stack.peek()) {
                    s3 = stack.pop();
                }
                stack.push(nums[i]);
            }
        }
        return false;
    }
}