LeetCode（154） Find Minimum in Rotated Sorted Array II

题目

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.

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分析

在含有重复元素的旋转有序序列中查找最小元素。

与上一题类似， LeetCode（153） Find Minimum in Rotated Sorted Array 153题中的旋转有序数组不包含重复元素，而此题允许重复元素，增加了一点难度。

我想题目重点考察的还是沿用二分查找的方法解决，思路参考。

AC代码

class Solution {public:    //方法一：使用stl库函数    int findMin1(vector<int>& nums) {        if (nums.empty())            return 0;        vector<int>::iterator iter = min_element(nums.begin(), nums.end());        return *iter;    }    //方法二：整个数列除一处为最大值到最小值的跳转外，为两部分的递增    int findMin2(vector<int>& nums)    {        if (nums.empty())            return 0;        if (nums.size() == 1)            return nums[0];        for (size_t i = 1; i < nums.size(); ++i)        {            if (nums[i - 1] > nums[i])                return nums[i];        }//for        //没有找到跳转元素，则序列无旋转        return nums[0];    }    int findMin(vector<int> &nums)    {        if (nums.empty())            return 0;        else if (nums.size() == 1)            return nums[0];        else{            int lhs = 0, rhs = nums.size() - 1;            while (lhs < rhs && nums[lhs] >= nums[rhs])            {                int mid = (lhs + rhs) / 2;                if (nums[lhs] > nums[mid])                    rhs = mid;                else if (nums[lhs] == nums[mid])                    ++lhs;                else                    lhs = mid + 1;            }//while            return nums[lhs];        }    }};

GitHub测试程序源码