README_EN.md

456. 132 Pattern

中文文档

Description

Given a sequence of n integers a₁, a₂, ..., a_n, 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: 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].

Solutions

Python3

class Solution:
    def find132pattern(self, nums: List[int]) -> bool:
        ak = float('-inf')
        stack = []
        for num in nums[::-1]:
            if num < ak:
                return True
            while stack and num > stack[-1]:
                ak = stack.pop()
            stack.append(num)
        return False

Java

class Solution {
    public boolean find132pattern(int[] nums) {
        int ak = Integer.MIN_VALUE;
        Deque<Integer> stack = new ArrayDeque<>();
        for (int i = nums.length - 1; i >= 0; --i) {
            if (nums[i] < ak) {
                return true;
            }
            while (!stack.isEmpty() && nums[i] > stack.peek()) {
                ak = stack.pop();
            }
            stack.push(nums[i]);
        }
        return false;
    }
}

...