Find Minimum in Rotated Sorted Array (minimum number of rotation of the array)

Subject description:

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.

　　There are original title directly on the code on this question "to prove safety offer"

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];
}

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

 

　　The above-described procedure does not consider the presence of the same element of the array. If we consider, then, the code needs to be modified.

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];
}

Finally, attach a solution on "to prove safety offer":

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：

　　Solution on the "wins the offer" is not necessarily the best! ! !

 

Reproduced in: https: //www.cnblogs.com/gattaca/p/4643750.html