leetcode456-132 mode

Given a sequence of integers: a _{. 1} , a ₂ , ..., a _n- , subsequence a a 132 mode _I , a _J , a _K is defined as: When I <J <K, a _I <a _K <a _J . Design an algorithm, when there is given a sequence of n numbers, to verify whether this sequence contains a subsequence 132 mode.

Note: the value of n is less than 15,000.

Example 1:

Input: [1, 2, 3, 4]
Output: False
Explanation: absence pattern sequence 132 sequences.

Example 2:

Input: [3, 1, 4, 2]
Output: True
Explanation: the sequence has a sequence pattern 132: [1, 4, 2].

Example 3:

Input: [-1, 3, 2, 0]
Output: True
Explanation: sequence has three sequences 132 modes: [-1, 3, 2], [-1, 3, 0] and [-1 , 2, 0].

Solution 2

2.1 ideas

1. Violence, traversing from the tail to the head, if the current element is the last element 132, respectively, the front and center must have less than a current element and an element is greater than the current element.

2. Stack, reference https://leetcode-cn.com/problems/132-pattern/solution/132mo-shi-by-leetcode-2/

2.2 Code

1. violence

class Solution {
    public boolean find132pattern(int[] nums) {
        for(int i = nums.length-1;i>=2;i--){
            int tail = nums[i];
            boolean hasPeek = false;
            for(int j = i-1;j>=0;j--){
                if(nums[j]>tail) hasPeek = true;
                else if(hasPeek&&nums[j]<tail) return true;
            }
        }
        return false;
    }
}

2. Stack

public class Solution {
    public boolean find132pattern(int[] nums) {
        if (nums.length <  大专栏  leetcode456-132模式3)
            return false;
        Stack<Integer> stack = new Stack<>();
        int[] min = new int[nums.length];
        min[0] = nums[0];
        for (int i = 1; i < nums.length; i++)
            min[i] = Math.min(min[i - 1], nums[i]);
        for (int j = nums.length - 1; j >= 0; j--) {
            if (nums[j] > min[j]) {
                while (!stack.isEmpty() && stack.peek() <= min[j])
                    stack.pop();
                if (!stack.isEmpty() && stack.peek() < nums[j])
                    return true;
                stack.push(nums[j]);
            }
        }
        return false;
    }
}