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).

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.

 

 to see which companies asked this question

1.做了下标映射

class Solution {public:        int binary_search(vector
      
        &nums, int target) {        int l = 0;        int r = nums.size() -1;        while (l<=r) {            int mid = l + ((r - l) >> 1);            if (nums[mid] == target) {                return mid;            } else if (nums[mid] < target) {                l = mid + 1;            } else {                r = mid - 1;            }        }        return -1;    }    int search(vector
       
        & nums, int target) {        //映射        int i = 0;        vector
        
          tmp_nums;        vector
         
           tmp_index;        int j = 0;        for (i=0; i