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];
}
}
};
【LeetCode】Find Minimum in Rotated Sorted Array II (2 solutions)
原文：http://www.cnblogs.com/ganganloveu/p/4081483.html