剑指 Offer 11. 旋转数组的最小数字-7月22日

class Solution {
public:
    int minArray(vector<int>& numbers) {
        int low = 0;
        int high = numbers.size()-1;
        int s;
        while(low!=high){
            s = low + (high - low)/2;
            if(numbers[high]>numbers[low]) return numbers[low];
            else if(numbers[high]<numbers[low]&&numbers[s]>=numbers[low]) low = s + 1;
            else if(numbers[high]<numbers[low]&&numbers[s]<=numbers[low]) high = s;
            else if(numbers[high]==numbers[low]){high--;}
        }
        return numbers[low];
    }
};
//二分查找
/*
先把两个指针low和high 指向数组首尾。若最小元素的指针（下标）是m，二分过程中的分界下标是s。
可能存在以下几种情况：
1.n[high]>n[low] return n[low]
2.n[high]<n[low] n[s]>=n[low] low = s+1
3.n[high]<n[low] n[s]<=n[high] high = s
4.n[high]==n[low] high--;//神来之笔，遇到无法判断二分到哪一边时，可以尝试直接把上边界象征性减小一点，重新判断
*/