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].
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。
示例1:
输入: [1, 2, 3, 4] 输出: False 解释: 序列中不存在132模式的子序列。
示例 2:
输入: [3, 1, 4, 2] 输出: True 解释: 序列中有 1 个132模式的子序列： [1, 4, 2].
示例 3:
输入: [-1, 3, 2, 0] 输出: True 解释: 序列中有 3 个132模式的的子序列: [-1, 3, 2], [-1, 3, 0] 和 [-1, 2, 0].
164ms
1 struct Stack<T> {
2 var items = [T]()
3 mutating func push(item: T) {
4 items.append(item)
5 }
6 mutating func pop() -> T {
7 return items.removeLast()
8 }
9 func top() -> T {
10 return items.last!
11 }
12 func empty() -> Bool {
13 return items.count == 0
14 }
15 }
16
17 class Solution {
18 func find132pattern(_ nums: [Int]) -> Bool {
19 var third = Int.min
20 var s = Stack<Int>()
21 for i in stride(from: nums.count - 1, through: 0, by: -1) {
22 if nums[i] < third {
23 return true
24 }else {
25 while !s.empty() && (nums[i] > s.top()) {
26 third = s.top()
27 _ = s.pop()
28 }
29 s.push(item: nums[i])
30 }
31 }
32 return false
33 }
34 }
324ms
1 class Solution {
2 func find132pattern(_ nums: [Int]) -> Bool {
3 var hanoi = Array<Int>.init();
4 var medNum = Int.min
5 for index in stride(from: nums.count-1, through: 0, by: -1) {
6 if nums[index] < medNum {
7 return true
8 } else {
9 while (hanoi.count > 0 && hanoi.last! < nums[index]) {
10 medNum = hanoi.last!
11 hanoi.removeLast()
12 }
13 }
14 hanoi.append(nums[index])
15 }
16 return false
17 }
18 }
1348ms
1 class Solution {
2 func find132pattern(_ nums: [Int]) -> Bool {
3 if nums.count < 3{
4 return false
5 }
6 var newNums = [nums[0]]
7 for i in 1..<nums.count {
8 if nums[i] != nums[i-1]{
9 newNums.append(nums[i]);
10 }
11 }
12 if newNums.count < 3{
13 return false
14 }
15 var minArr = [newNums[0]]
16 var min = newNums[0]
17 for i in newNums{
18 if i < min{
19 min = i
20 }
21 minArr.append(min)
22 }
23 for i in 1..<newNums.count-1{
24 let a = minArr[i]
25 let b = newNums[i]
26 if a >= b{
27 continue
28 }
29 for j in (i+1)..<newNums.count{
30 let c = newNums[j]
31 if c > a && c < b {
32 return true
33 }
34 }
35 }
36 return false
37 }
38 }
4388ms
1 class Solution {
2 func find132pattern(_ nums: [Int]) -> Bool {
3 if nums.count < 3 {
4 return false
5 }
6 var i = nums.count
7 var min = nums[i - 1]
8 while (i > 0) {
9 min = nums[i - 1]
10 var j = nums.count
11 var max = min
12 while (j > i) {
13 let t = nums[j - 1]
14 if t > min {
15 if max > min {
16 if t > max {
17 return true
18 }else {
19 max = t
20 }
21 }else {
22 max = t
23 }
24 }
25 j = j - 1
26 }
27 i = i - 1
28 }
29 return false
30 }
31 }
6656ms
1 class Solution {
2 func find132pattern(_ nums: [Int]) -> Bool {
3 var n:Int = nums.count
4 var i:Int = 0
5 var j:Int = 0
6 var k:Int = 0
7 while (i < n) {
8 while (i < n - 1 && nums[i] >= nums[i + 1])
9 {
10 i += 1
11 }
12 j = i + 1;
13 while (j < n - 1 && nums[j] <= nums[j + 1])
14 {
15 j += 1
16 }
17 k = j + 1
18 while (k < n) {
19 if nums[k] > nums[i] && nums[k] < nums[j]
20 {
21 return true
22 }
23 k += 1
24 }
25 i = j + 1
26 }
27 return false
28 }
29 }
原文地址：https://www.cnblogs.com/strengthen/p/10343200.html
时间： 2024-07-30 12:57:28