Find Minimum in Rotated Sorted Array(旋转数组的最小数字)

题目描述：

Suppose a sorted array is rotated at some pivot unknown to you beforehand.

(i.e.,        might become       ).

Find the minimum element.

You may assume no duplicate exists in the array.

　　这道题《剑指offer》上有原题，直接上代码

solution：

int findMin(vector<int>& nums) {
    int start = ;
    int end = nums.size() - ;
    while (start < end)
    {
        if (nums[start] < nums[end])
            break;
        int mid = start + (end - start) / ;
        if (nums[mid] >= nums[start])
            start = mid + ;
        else
            end = mid;
    }
    return nums[start];
}

参考：https://leetcode.com/discuss/13389/compact-and-clean-c-solution

　　上述程序没有考虑数组中存在相同元素这一情况。如果考虑的话，代码需要修改。

solution：

int findMin(vector<int>& nums) {
    int start = ;
    int end = nums.size() - ;
    while (start < end)
    {
        if (nums[start] < nums[end])
            break;
        int mid = start + (end - start) / ;
        if (nums[mid] > nums[end])
            start = mid + ;
        else if (nums[mid] < nums[end])
            end = mid;
        else
        {
            ++start;
            --end;
        }
    }
    return nums[start];
}

最后附上一个《剑指offer》上的解法：

int MinInOrder(vector<int> &nums, int start, int end)
{
    int min = nums[start];
    for (int i = ; i < nums.size(); ++i)
    {
        if (nums[i] < min)
            min = nums[i];
    }
    return min;
}

int findMin(vector<int>& nums) {
    int start = ;
    int end = nums.size() - ;
    while (start < end)
    {
        if (nums[start] < nums[end])
            break;
        int mid = start + (end - start) / ;

        if (nums[mid] == nums[start] && nums[mid] == nums[end])
        {
            return MinInOrder(nums, start, end);
        }
        if (nums[mid] >= nums[start])
            start = mid + ;
        else
            end = mid;
    }
    return nums[start];
}

PS：

　　《剑指offer》上的解法不一定是最好的！！！