0456.132 Pattern

456. 132 Pattern #

题目 #

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].

题目大意 #

给定一个整数序列:a1, a2, …, an,一个 132 模式的子序列 ai, aj, ak 被定义为:当 i < j < k 时,ai < ak < aj。设计一个算法,当给定有 n 个数字的序列时,验证这个序列中是否含有 132 模式的子序列。注意:n 的值小于 15000。

解题思路 #

  • 这一题用暴力解法一定超时
  • 这一题算是单调栈的经典解法,可以考虑从数组末尾开始往前扫,维护一个递减序列

代码 #


package leetcode

import (
	"fmt"
	"math"
)

// 解法一 单调栈
func find132pattern(nums []int) bool {
	if len(nums) < 3 {
		return false
	}
	num3, stack := math.MinInt64, []int{}
	for i := len(nums) - 1; i >= 0; i-- {
		if nums[i] < num3 {
			return true
		}
		for len(stack) != 0 && nums[i] > stack[len(stack)-1] {
			num3 = stack[len(stack)-1]
			stack = stack[:len(stack)-1]
		}
		stack = append(stack, nums[i])
		fmt.Printf("stack = %v \n", stack)
	}
	return false
}

// 解法二 暴力解法,超时!
func find132pattern1(nums []int) bool {
	if len(nums) < 3 {
		return false
	}
	for j := 0; j < len(nums); j++ {
		stack := []int{}
		for i := j; i < len(nums); i++ {
			if len(stack) == 0 || (len(stack) > 0 && nums[i] > nums[stack[len(stack)-1]]) {
				stack = append(stack, i)
			} else if nums[i] < nums[stack[len(stack)-1]] {
				index := len(stack) - 1
				for ; index >= 0; index-- {
					if nums[stack[index]] < nums[i] {
						return true
					}
				}
			}
		}
	}
	return false
}


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