Find Minimum in Rotated Sorted Array II

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.

http://www.lintcode.com/en/problem/find-minimum-in-rotated-sorted-array-ii/

Discussion

存在重复元素，那么跟Find Minimum in Rotated Sorted Array问题相比，if(num[mid] == num[end]) end--;即可。注意不是start++，因为会跳过最小值了。

这个题跟Search in Rotated Sorted Array 问题扩展到Search in Rotated Sorted Array II问题的思路是一样的。

Solution

class Solution {
public:
    /**
     * @param num: the rotated sorted array
     * @return: the minimum number in the array
     */
    int findMin(vector<int> &num) {
        // write your code here
        int start = 0;
        int end = num.size() -1;
        while(start + 1 < end) {
            int mid = start+ (end-start)/2;
            if(num[mid] < num[end]) {
                end = mid;
            } else if(num[mid] > num[end]) {
                start = mid;
            } else {
                end--;
            }
        }
        if(num[start]<=num[end]) return num[start];
        return num[end];
    }
};