[Swift]LeetCode154. 尋找旋轉排序數組中的最小值 II | Find Minimum in Rotated Sorted Array II

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Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.

(i.e.,  [0,1,2,4,5,6,7] might become  [4,5,6,7,0,1,2]).

Find the minimum element.

The array may contain duplicates.

Example 1:

Input: [1,3,5]
Output: 1

Example 2:

Input: [2,2,2,0,1]
Output: 0

Note:

• This is a follow up problem to Find Minimum in Rotated Sorted Array.
• Would allow duplicates affect the run-time complexity? How and why?

---

假設按照升序排序的數組在預先未知的某個點上進行了旋轉。

( 例如，數組 [0,1,2,4,5,6,7] 可能變爲 [4,5,6,7,0,1,2] )。

請找出其中最小的元素。

注意數組中可能存在重複的元素。

示例 1：

輸入: [1,3,5]
輸出: 1

示例 2：

輸入: [2,2,2,0,1]
輸出: 0

說明：

• 這道題是 尋找旋轉排序數組中的最小值 的延伸題目。
• 容許重複會影響算法的時間複雜度嗎？會如何影響，爲何？

---

 

16ms

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

---

52ms

 1 class Solution {
 2     func findMin(_ nums: [Int]) -> Int {
 3         let arraySize = nums.count
 4 
 5         if arraySize == 1 { return nums[0] }
 6         if arraySize == 2 { return min(nums[0], nums[1]) }
 7         let midElement = arraySize / 2
 8         if nums[midElement] > nums[arraySize-1] {
 9             return findMin(Array(nums[midElement...arraySize-1]))
10         } else if nums[midElement] < nums[0] {
11             return findMin(Array(nums[0...midElement]))
12         } else {
13             return min(findMin(Array(nums[midElement...arraySize-1])), findMin(Array(nums[0...midElement])))
14         }
15     }
16 }

---

60ms

 1 class Solution {
 2     func findMin(_ nums: [Int]) -> Int {
 3 
 4         if nums.isEmpty {
 5             return -1
 6         }
 7 
 8         if nums.count == 1 {
 9             return nums[0]
10         }
11         
12         var beigin = nums[0]
13         var i = 1
14         while i < nums.count {
15             if nums[i] >= beigin {
16                 beigin = nums[i]
17                 i += 1
18             } else {
19                 return nums[i]
20             }
21         }
22 
23         return nums[0]
24     }
25 }

---

76ms

1 class Solution {
2     func findMin(_ nums: [Int]) -> Int {
3         guard nums.count > 0 else { return -1 }
4         return nums.min()!
5     }
6 }