# 456 132 Pattern

## 456. [132 Pattern](https://leetcode.com/problems/132-pattern/description/)

## 1. Question

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

**Note:**&#x6E; 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].
```

## 2. Implementation

思路: 这题只要返回true or false，所以我们从后往前遍历数组，利用单调递减stack寻找第二大的数，一旦遍历的过程中找到一个前面的数小于第二大的数，则我们找到pattern

```java
class Solution {
    public boolean find132pattern(int[] nums) {
        if (nums == null || nums.length == 0) {
            return false;
        }

        int secondMax = Integer.MIN_VALUE;
        Stack<Integer> stack = new Stack<>();

        for (int i = nums.length - 1; i >= 0; i--) {
            if (nums[i] < secondMax) {
                return true;
            }
            else {
                while (!stack.isEmpty() && stack.peek() < nums[i]) {
                    secondMax = stack.pop();
                }
                stack.push(nums[i]);
            }
        }
        return false;
    }
}
```

## 3. Time & Space Complexity

时间和空间复杂度都是O(n)


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