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