Find Minimum in Rotated Sorted Array 2 寻找旋转有序数组的最小值之二

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.

 

这道寻找旋转有序重复数组的最小值是对之前那道问题的延伸，当数组中存在大量的重复数字时，就会破坏二分查找法的机制，我们无法取得O(lgn)的时间复杂度，又将会回到简单粗暴的O(n)，比如如下两种情况：

{2, 2, 2, 2, 2, 2, 2, 2, 0, 1, 1, 2} 和 {2, 2, 2, 0, 2, 2, 2, 2, 2, 2, 2, 2}， 我们发现，当第一个数字和最后一个数字，还有中间那个数字全部相等的时候，二分查找法就崩溃了，因为它无法判断到底该去左半边还是右半边。这种情况下，我们只能回到最原始的那种简单粗暴的方法，把整个数组从头到尾撸一遍，找出最小的那个。代码如下：

 

class Solution {public:    int findMin(vector<int> &num) {if (num.size() <= 0) return 0;if (num.size() == 1) return num[0];        int left = 0, right = num.size() - 1;if (num[left] > num[right]) {while (left != (right - 1)) {int mid = (left + right) / 2;if (num[left] <= num[mid]) left = mid;else right = mid;}return min(num[left], num[right]);} else if (num[left] == num[right]) {int res = num[0];for (int i = 1; i < num.size(); ++i) {res = min(res, num[i]);}return res;}return num[0];    }};