描述

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.

分析

使用二分查找，注意边界条件

代码实现 c++
class solution{
pubic: 
    int search(const vector<int>& nums,int target){
        int first = 0,last = nums.size();
        while(first != last){
            const int mid = first + (last - first) / 2;
            if (nums[mid] == target)
                return mid;
            if (nums[first] <= nums[mid]){
                // 注意此处有等于号，表明在闭区间内
                if(nums[first] <= target && target < nums[mid])
                    last = mid;
                 else
                    first = mid + 1;
             } else {
                    if(nums[mid] < target && target <= nums[last - 1])
                        first = mid + 1;
                    else
                        last = mid;
              }
          }
          return -1;
    }
};

代码实现 python
class Solution(object):
    def search(self, nums, target):
       if not nums:
            return -1
            
        start, end = 0, len(nums) - 1
        while start + 1 < end:
            mid = int(start + (end - start) / 2)
            if nums[mid] == target:
                return mid
            elif nums[mid] > nums[start]:
                if target >= nums[start] and target <= nums[mid]:
                    end = mid
                else:
                    start = mid
            else:
                if target >= nums[mid] and target <= nums[end]:
                    start = mid
                else:
                    end = mid
                    
        if nums[start] == target:
            return start
        if nums[end] == target:
            return end
        return -1