456.132-Pattern.md

456. 132 Pattern

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题目地址

https://leetcode.com/problems/132-pattern/

题目描述

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].

代码

Approach 1: Brute Force

The simplest solution is to consider every triplet (i, j, k) and check if the corresponding numbers satisfy the 132 criteria. If any such triplet is found, we can return a True value. If no such triplet is found, we need to return a False value.

Complexity Analysis

• Time complexity : O(n^3). Three loops are used to consider every possible triplet. Here, n_n_ refers to the size of nums array.
• Space complexity : O(1). Constant extra space is used.

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

    return false;
  }
}

Approach #2 Better Brute Force

Complexity Analysis

• Time complexity : O(n^2). Two loops are used to find the nums[j],nums[k] pairs. Here, n refers to the size of nums array.
• Space complexity : O(1). Constant extra space is used.

class Solution {
  public boolean find132pattern(int[] nums) {
    int min_i = Integer.MAX_VALUE;
    for (int j = 0; j < nums.length - 1; j++) {
      min_i = Math.min(min_i, nums[j]);
      for (int k = j + 1; k < nums.length; k++) {
        if (nums[k] < nums[j] && min_i < nums[k]) {
          return true;
        }
      }
    }

    return false;
  }
}

Approach #3 Stack

单调递减栈

从左到右，生成min[i]左边最小的数组

从右到左，找出 num[j] > min[j] && stack.peek() < num[j] && stack.peek() > min[j]

class Solution {
  public boolean find132pattern(int[] nums) {
    if (nums.length < 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;
  }
}

Approach 4: Using Array as a Stack

public class Solution {
    public boolean find132pattern(int[] nums) {
        if (nums.length < 3)
            return false;
        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, k = nums.length; j >= 0; j--) {
            if (nums[j] > min[j]) {
                while (k < nums.length && nums[k] <= min[j])
                    k++;
                if (k < nums.length && nums[k] < nums[j])
                    return true;
                nums[--k] = nums[j];
            }
        }
        return false;
    }
}