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., 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.

　　这道题《剑指offer》上有原题，直接上代码

solution：

int findMin(vector<int>& nums) {
    int start = 0;
    int end = nums.size() - 1;
    while (start < end)
    {
        if (nums[start] < nums[end])
            break;
        int mid = start + (end - start) / 2;
        if (nums[mid] >= nums[start])
            start = mid + 1;
        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 = 0;
    int end = nums.size() - 1;
    while (start < end)
    {
        if (nums[start] < nums[end])
            break;
        int mid = start + (end - start) / 2;
        if (nums[mid] > nums[end])
            start = mid + 1;
        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 = 1; i < nums.size(); ++i)
    {
        if (nums[i] < min)
            min = nums[i];
    }
    return min;
}

int findMin(vector<int>& nums) {
    int start = 0;
    int end = nums.size() - 1;
    while (start < end)
    {
        if (nums[start] < nums[end])
            break;
        int mid = start + (end - start) / 2;

        if (nums[mid] == nums[start] && nums[mid] == nums[end])
        {
            return MinInOrder(nums, start, end);
        }
        if (nums[mid] >= nums[start])
            start = mid + 1;
        else
            end = mid;
    }
    return nums[start];
}

PS：

　　《剑指offer》上的解法不一定是最好的！！！