Leetcode | Find Minimum in Rotated Sorted Array I && II-优快云博客

本文介绍了一种在未知旋转点的有序数组中查找最小元素的方法，并提供了两种情况下的C++实现：一种是没有重复元素的情况，另一种是允许存在重复元素的情况。通过二分查找法高效解决问题。

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Suppose a sorted array 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.

You may assume no duplicate exists in the array.

二分查找就行了。

 1 class Solution {
 2 public:
 3     int findMin(vector<int> &arr) {
 4         if (arr.empty()) return -1;
 5         int l = 0, h = arr.size() - 1, m, min = INT_MAX;
 6         while (l <= h) {
 7             m = l + (h - l) / 2;
 8             if (arr[m] >= arr[l]) {
 9                 if (min > arr[l]) min = arr[l];
10                 l = m + 1;
11             } else {
12                 if (arr[m] < min) min = arr[m];
13                 h = m - 1;
14             }
15         }
16         return min;
17     }
18 };

Follow up for "Find Minimum in Rotated Sorted Array":
What if duplicates are allowed?

Would this affect the run-time complexity? How and why?
Suppose a sorted array 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.

 1 class Solution {
 2 public:
 3     int findMin(vector<int> &arr) {
 4         if (arr.empty()) return -1;
 5         int l = 0, h = arr.size() - 1, m, min = INT_MAX;
 6         while (l <= h) {
 7             m = l + (h - l) / 2;
 8             if (arr[m] > arr[l]) {
 9                 if (min > arr[l]) min = arr[l];
10                 l = m + 1;
11             } else if (arr[m] < arr[l]) {
12                 if (arr[m] < min) min = arr[m];
13                 h = m - 1;
14             } else {
15                 if (arr[m] < min) min = arr[m];
16                 l++;
17             }
18         }
19         return min;
20     }
21 };