地址

https://leetcode.com/problems/find-minimum-in-rotated-sorted-array-ii/

题目

Suppose an array sorted in ascending order 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.

Example 1:

Input: [1,3,5]
Output: 1

Example 2:

Input: [2,2,2,0,1]
Output: 0

Note:

• This is a follow up problem to Find Minimum in Rotated Sorted Array.
• Would allow duplicates affect the run-time complexity? How and why?

思路

对于rotate后的数组，里面的元素大概长这样[5,5,5,6,7,...,1,2,3,4,5,5,5,5]（数组内可能有重复元素），所以数组中最小元素a[i]显然是第一个大于a[0]的数。

可以看出，a[i]左边的元素都大于等于a[0]，a[i]及a[i]右边的元素都小于等于a[0]。所以我们可以通过二分查找i。

但是由于a[i]的左右两边都存在可能等于a[0]的元素，二分时不好区分，所以要先处理掉数组最右边和a[0]相等的元素（也就是修改二分的上界）。

代码

class Solution {
public:
    int findMin(vector<int>& a) {
        if(a.size()==1) return a[0];
        int l=1,r=a.size()-1;
        while(r>1&&a[r]==a[0]) --r;
        if(r==0||a[r] > a[0]) return a[0];
        while(l<r)
        {
            int mid=l+r>>1;
            if(check(mid, a))
                l=mid+1;
            else
                r=mid;
        }
        return a[r];
    }
    
    bool check(int x, vector<int>& a)
    {
        return a[x]>=a[0];
    }
};