【LeetCode】Find Minimum in Rotated Sorted Array II (2 solutions)

Find Minimum in Rotated Sorted Array II

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.

解法一：暴力解法，直接使用algorithm库中的求最小元素函数

需要遍历整个vector

class Solution {
public:
    int findMin(vector<int> &num) {
        if(num.empty())
            return 0;
        vector<int>::iterator iter = min_element(num.begin(), num.end());
        return *iter;
    }
};

解法二：利用sorted这个信息。如果平移过，则会出现一个gap，也就是从最大元素到最小元素的跳转。如果没有跳转，则说明没有平移。

比上个解法可以省掉不少时间，平均情况下不用遍历vector了。

class Solution {
public:
    int findMin(vector<int> &num) {
        if(num.empty())
            return 0;
        else if(num.size() == 1)
            return num[0];
        else
        {
            for(vector<int>::size_type st = 1; st < num.size(); st ++)
            {
                if(num[st-1] > num[st])
                    return num[st];
            }
            return num[0];
        }
    }
};