Leetcode 456 - 132 Pattern

题目：

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].

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思路：

该题利用到了栈这一数据结构。该题求解的是132 Pattern。栈中存储当前最大的数。同时设置一个值third。用于存储132中3位置的数。将给定的数组从后往前遍历，一旦当前数据大于栈中的数据就弹出栈。third存储弹出数据中最大的值。一旦当前遍历到的数据小于third则返回true。此时third记录着3位置的数据，栈中记录着2位置的数据，遍历到的数据为1位置上的数据。

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代码：

public boolean find132pattern(int[] nums) {
        if(nums == null || nums.length == 0)
            return false;
        boolean result = false;
        int third = Integer.MIN_VALUE;
        Stackstack = new Stack();
        for(int i=nums.length-1;i>=0;i--){
            if(nums[i] < third)
                return true;
            while(!stack.isEmpty() && nums[i] > stack.peek()){
                third = Math.max(third, stack.pop());
            }
            stack.push(nums[i]);
        }
        return result;
}