#160 Find Minimum in Rotated Sorted Array II

题目描述：

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.

 Notice

The array may contain duplicates.

Have you met this question in a real interview? 

Yes

Example

Given [4,4,5,6,7,0,1,2] return 0.

题目思路：

这题和#159基本上一样，只是为了省去repeat number的考虑，加了俩while循环skip掉重复的numbers。

Mycode（AC = 14ms）：

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
        if (num.size() == 1) {
            return num[0];
        }
        
        int l = 0, r = num.size() - 1;
        while (l + 1 < r) {
            int mid = (l + r) / 2;
            
            // skip the repeated numbers
            while (l + 1 < r && num[l] == num[l + 1]) {
                l++;
            }
            if (l + 1 >= r) break;
            
            // skip the repeated numbers
            while (l + 1 < r && num[r] == num[r - 1]) {
                r--;
            }
            if (l + 1 >= r) break;
            
            if (num[mid] < num[l] || num[mid] < num[r]) {
                r = mid;
            }
            else {
                l = mid;
            }
            
        }
        
        return num[l] < num[r] ? num[l] : num[r];
    }
};