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.

O(n)的方法

找到有序的一边。然后取min就可以。

关键就是找到rotate list有序的那一边

循环实现：

class Solution {
public:
    int findMin(vector<int> &num) {
        return find(num,0,num.size()-1);
    }
    int find(vector<int> &num,int left,int right){
        if(left==right)
            return num[left];
        int mi = INT_MAX;
        while(left<=right){
            int mid = (left+right)/2;
            int i=mid,j=mid;
            while(i>=left && num[i]==num[mid])
                i--;
            while(j<=right && num[j]==num[mid])
                j++;
            mi = min(mi,num[mid]);
            if(i<left){
                if(j>right){
                    break;
                }else
                    left = j;
            }else{
                if(j>right)
                    right = i;
                else{
                    if(num[left]<=num[i]){
                        mi = min(num[left],mi);
                        left = j;
                    }else{
                        mi = min(num[j],mi);
                        right = i;
                    }
                }
            }
        }
        return mi;
    }
};

递归实现：

class Solution {
public:
    int findMin(vector<int> &num) {
        return find(num,0,num.size()-1);
    }
    int find(vector<int> &num,int left,int right){
        if(left==right)
            return num[left];
        
            int mid = (left+right)/2;
            int i=mid,j=mid;
            while(i>=left && num[i]==num[mid])
                i--;
            while(j<=right && num[j]==num[mid])
                j++;
            if(i<left){
                if(j>right)
                    return num[mid];
                return min(num[mid],find(num,j,right));
            }else{
                if(j>right)
                    return min(num[mid],find(num,left,i));
                if(num[left]<=num[i]){
                    return min(num[mid],min(num[left],find(num,j,right)));
                }else{
                    return min(num[mid],min(num[j],find(num,left,i)));
                }
            }
        
    }
};