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103 lines
2.3 KiB
Markdown
Executable File
103 lines
2.3 KiB
Markdown
Executable File
# [456. 132 Pattern](https://leetcode.com/problems/132-pattern/)
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## 题目
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Given a sequence of n integers a1, a2, ..., an, 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.
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**Note**: n will be less than 15,000.
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**Example 1**:
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Input: [1, 2, 3, 4]
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Output: False
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Explanation: There is no 132 pattern in the sequence.
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**Example 2**:
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Input: [3, 1, 4, 2]
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Output: True
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Explanation: There is a 132 pattern in the sequence: [1, 4, 2].
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**Example 3**:
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Input: [-1, 3, 2, 0]
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Output: True
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Explanation: There are three 132 patterns in the sequence: [-1, 3, 2], [-1, 3, 0] and [-1, 2, 0].
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## 题目大意
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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。
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## 解题思路
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- 这一题用暴力解法一定超时
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- 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
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## 代码
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```go
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package leetcode
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import (
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"fmt"
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"math"
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)
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// 解法一 单调栈
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func find132pattern(nums []int) bool {
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if len(nums) < 3 {
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return false
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}
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num3, stack := math.MinInt64, []int{}
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for i := len(nums) - 1; i >= 0; i-- {
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if nums[i] < num3 {
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return true
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}
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for len(stack) != 0 && nums[i] > stack[len(stack)-1] {
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num3 = stack[len(stack)-1]
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stack = stack[:len(stack)-1]
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}
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stack = append(stack, nums[i])
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fmt.Printf("stack = %v \n", stack)
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}
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return false
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}
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// 解法二 暴力解法，超时！
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func find132pattern1(nums []int) bool {
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if len(nums) < 3 {
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return false
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}
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for j := 0; j < len(nums); j++ {
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stack := []int{}
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for i := j; i < len(nums); i++ {
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if len(stack) == 0 || (len(stack) > 0 && nums[i] > nums[stack[len(stack)-1]]) {
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stack = append(stack, i)
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} else if nums[i] < nums[stack[len(stack)-1]] {
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index := len(stack) - 1
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for ; index >= 0; index-- {
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if nums[stack[index]] < nums[i] {
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return true
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}
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}
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}
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}
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}
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return false
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}
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``` |