456 132 Pattern

456. 132 Pattern

1. Question

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].

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

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)