Leetcode|Find Minimum in Rotated Sorted Array II(有重复元素的二分查找）

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1：

//自己想的，遇到首尾和mid相等的，检验mid是左边还是在右边，检验时只能挨个查找。
int findMin1(vector<int>& nums) {
int n=nums.size();
if(n==1) return nums[0];
int first=0,last=n-1;
while(first<last){
int mid=(first+last)/2;
if(nums[last]<nums[mid]) first=mid+1;
else{
if(nums[first]==nums[mid]&&nums[mid]==nums[last]){
int l=mid;
while(l<last&&nums[++l]==nums[last]);//检验mid是否在右边
if(l==last) last=mid;
else first=mid+1;
}
else last=mid;
}
}
return nums[first];
}

解法2：思路更加清晰，易懂。

//按照陈老师的优化了思路，直接把开始的first(nums[first]==nums[last])的重复的去掉。
//后面和没重复的一样了！
int findMin(vector<int>& nums) {
int n=nums.size();
if(n==1) return nums[0];
int first=0,last=n-1;
while(first<last){
while(nums[first]==nums[last]&&first<last){
first++;
}
int mid=(first+last)/2;
if(nums[last]<nums[mid]) first=mid+1;
else  last=mid;
}
return nums[first];
}