LeetCode-33. Search in Rotated Sorted Array

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本文介绍了一种高效的算法来解决LeetCode第33题搜索旋转排序数组的问题。通过两次二分查找，先确定旋转点，再找到目标值的位置，确保了时间复杂度为O(log n)。

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LeetCode-33. Search in Rotated Sorted Array

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.

Your algorithm's runtime complexity must be in the order of O(log n).

Example 1:

Input: nums = [4,5,6,7,0,1,2], target = 0
Output: 4

Example 2:

Input: nums = [4,5,6,7,0,1,2], target = 3
Output: -1

解题思路：本题要求找出某一个target在一个按顺序排序然后旋转之后的数组中的位置，分两步进行，第一步就是利用二分查找的方法找出数组的最小位置，记为start，第二步就是利用二分查找的方法查找target，注意realmid=mid+start 才是按顺序排序的数组的真正中点。

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