[Swift]LeetCode162. 寻找峰值 | Find Peak Element

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A peak element is an element that is greater than its neighbors.

Given an input array nums, where nums[i] ≠ nums[i+1], find a peak element and return its index.

The array may contain multiple peaks, in that case return the index to any one of the peaks is fine.

You may imagine that nums[-1] = nums[n] = -∞.

Example 1:

Input: nums = 
Output: 2
Explanation: 3 is a peak element and your function should return the index number 2.[1,2,3,1]

Example 2:

Input: nums = 1,2,1,3,5,6,4]
Output: 1 or 5 
Explanation: Your function can return either index number 1 where the peak element is 2, 
             or index number 5 where the peak element is 6.
[

Note:

Your solution should be in logarithmic complexity.

---

峰值元素是指其值大于左右相邻值的元素。

给定一个输入数组 nums，其中 nums[i] ≠ nums[i+1]，找到峰值元素并返回其索引。

数组可能包含多个峰值，在这种状况下，返回任何一个峰值所在位置便可。

你能够假设 nums[-1] = nums[n] = -∞。

示例 1:

输入: nums = 
输出: 2
解释: 3 是峰值元素，你的函数应该返回其索引 2。[1,2,3,1]

示例 2:

输入: nums = 1,2,1,3,5,6,4]
输出: 1 或 5 
解释: 你的函数能够返回索引 1，其峰值元素为 2；
     或者返回索引 5， 其峰值元素为 6。
[

说明:

你的解法应该是 O(logN) 时间复杂度的。

---

32ms

 1 class Solution {
 2     func findPeakElement(_ nums: [Int]) -> Int {
 3         var low = 0
 4         var high = nums.count - 1
 5         
 6         while low != high {
 7             let mid = (low + high)/2
 8             
 9             if !nums.indices.contains(mid - 1) {
10                 if nums[mid + 1] < nums[mid] { return mid }
11                 low = mid + 1
12                 continue
13             }
14             
15             if nums[mid - 1] < nums[mid] && nums[mid + 1] < nums[mid] {
16                 return mid
17             } 
18             
19             if nums[mid - 1] < nums[mid] && nums[mid + 1] > nums[mid] {
20                 low = mid + 1
21                 continue
22             }
23             
24             high = mid - 1
25         }
26         
27         return high
28     }
29 }

---

36ms

 1 class Solution {
 2     func findPeakElement(_ nums: [Int]) -> Int {
 3             guard nums.count > 1 else{
 4                 return 0
 5             }
 6             var left = 0
 7             var right = nums.count  - 1
 8             while left < right{
 9                 let middle = (left + right) / 2
10                 if nums[middle] > nums[middle + 1] {
11                     right = middle 
12                 }else{
13                     left = middle + 1
14                 }
15             }
16         
17         return left
18         
19     }
20     
21 }

---

40ms

 1 class Solution {
 2     func findPeakElement(_ nums: [Int]) -> Int {
 3         if nums.count >= 3 {
 4             for i in Array(1..<nums.count - 1) {
 5                 let left = nums[i - 1]
 6                 let value = nums[i]
 7                 let right = nums[i + 1]
 8 
 9                 if value > left && value > right {
10                     return i
11                 }
12             }
13         }
14         
15         let lower = 0
16         let upper = nums.count - 1
17         
18         return nums[lower] >= nums[upper] ? lower : upper
19     }
20 }

---

44ms

 1 class Solution {
 2     func findPeakElement(_ nums: [Int]) -> Int {
 3         var low = 0
 4         var high = nums.count - 1
 5         
 6         while low != high {
 7             let mid = (low + high)/2
 8             
 9             if !nums.indices.contains(mid - 1) || nums[mid - 1] < nums[mid] {
10                 if nums[mid + 1] < nums[mid] { return mid }
11                 low = mid + 1
12                 continue
13             }
14             
15             high = mid - 1
16         }
17         
18         return high
19     }
20 }

---

52ms

 1 class Solution {
 2     func findPeakElement(_ nums: [Int]) -> Int {
 3 var left = 0
 4     var right = nums.count - 1
 5     var mid = 0
 6     
 7     while left < right {
 8         mid = (right - left) / 2 + left
 9         
10         if nums[mid] > nums[mid + 1] {
11             right = mid
12         } else {
13             left = mid + 1
14         }
15     }
16     return left
17     }
18 }