33. Search in Rotated Sorted Array 旋转排序数组极值的二分求法

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).

You are given a target value to search. If found in the array return its index, otherwise return -1.

You may assume no duplicate exists in the array.

class Solution {public:    int search(vector<int>& nums, int target) {        int n = nums.size();        int lo = 0, hi = n - 1;        while(lo < hi){//find smallest            int mid = (lo + hi) / 2;            if(nums[mid] > nums[hi]) lo = mid + 1;// 这里 nums[mid] > nums[hi] 一定大于 所以要 +1            else hi = mid;// (lo + hi) / 2 会使mid 一定小于hi 已经至少减1了 无需 hi = mid;        }        int L, R;        if(lo == 0){//如果n == 1 ，lo必等于0，必进这个分支            L = 0;            R = n - 1;        }else if(target >= nums[0]){            L = 0;            R = lo - 1;        }else{            L = lo;            R = n - 1;        }        while(L <= R){            if(target > nums[R] || target < nums[L] ) return -1;            int mid = (L + R) / 2;            if(nums[mid] == target) return mid;            else if(nums[mid] > target) R = mid - 1;//类似于第8行 一定要 -1            else L = mid + 1;        }                return L;            }};