[leetcode] 33. Search 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).

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.

解法一：

二分法，先确定那边是ordered数组。

class Solution {public:    int search(vector<int>& nums, int target) {        if(nums.empty()) return 0;        int left = 0, right = nums.size()-1;                while(left<=right){            int mid = (left+right)/2;            if(target== nums[mid]) return mid;            else if(nums[mid]<nums[right]){                if(target>nums[mid] && target<=nums[right]) left = mid +1;                else right = mid -1;                            } else{                if(target>=nums[left] && target < nums[mid]) right = mid -1;                else left = mid + 1;            }        }        return -1;            }};