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

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於左右相鄰值的元素。

給定一個輸入數組 nums，其中 nums[i] ≠ nums[i+1]，找到峰值元素並返回其索引。

數組可能包含多個峰值，在這種情況下，返回任何一個峰值所在位置即可。

你可以假設 nums[-1] = nums[n] = -∞。

示例 1:

輸入: nums = [1,2,3,1]
輸出: 2
解釋: 3 是峰值元素，你的函數應該返回其索引 2。

示例 2:

輸入: nums = [1,2,1,3,5,6,4]
輸出: 1 或 5 
解釋: 你的函數可以返回索引 1，其峰值元素為 2；
     或者返回索引 5， 其峰值元素為 6。

說明:

你的解法應該是 O(logN) 時間復雜度的。

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 }

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