Find Minimum in Rotated Sorted Array

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.

You may assume no duplicate exists in the array.

思路：二分缩小范围，直到找到。

注意: after rotating at pivot, cases may be (0 1 2 3) (3 2 1 0) (2 3 0 1) 

so think twice when you code !

class Solution {
public:
    int findMin(vector<int> &num) {
        int n = num.size();
        return find(num,0,n-1);
    }
    int find(vector<int> &num, int a, int b){
        int k = (a+b)/2;
        if(num[k]>num[b]) return find(num,k+1,b);
        else if(num[k]<num[a]) return find(num,a,k);
        else return num[a];
    }
};

另一种解法如下，单丝while(a<=b)就会出错。这是为何？？待我想想

class Solution {
public:
    int findMin(vector<int> &num) {
        int a=0;
        int b=num.size()-1;
        int k=0;
        while(a<b){
          k = (a+b)/2;
          if(num[k]>num[b]) a=k+1;
          else b=k;
      }
      return num[a];
    }
    
};

PS：sort by pivot 的方法是 插入排序。具体方法~~~