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website/content/ChapterFour/0456.132-Pattern.md

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+# [456. 132 Pattern](https://leetcode.com/problems/132-pattern/)
+
+
+## 题目
+
+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.
+
+**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
+
+```go
+
+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
+}
+
+```