LeetCode——033

/*
33. Search in Rotated Sorted Array My Submissions QuestionEditorial Solution
Total Accepted: 98590 Total Submissions: 326117 Difficulty: Hard
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.

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*/

/*
解题思路：
旋转数组的元素查找，依然采用折半查找
比非旋转数组多了一层判断

if(右半边有序）{
if(target在右半边）{
left=mid+1;
}
else{
right=mid-1;
}
//左半边有序
}else{
if（target在左半边）{
right=mid-1;
}
else{
left=mid+1;
}
}

*/

class Solution {
public:
    int search(vector<int>& nums, int target) {


        //旋转数组中查找数字，使用折半查找法

        if(nums.empty())return -1;
        int left=0,right=nums.size()-1;
        int mid;
        while(left<=right){

            mid=left+(right-left)/2;
            if(nums[mid]==target)return mid;
            //右半边有序
            else if(nums[mid]<nums[right]){

                if(nums[mid]<target&&nums[right]>=target)left=mid+1;
                else right=mid-1;
            //左半边有序
            }else{
                if(nums[mid]>target&& nums[left]<=target)right=mid-1;
                else left=mid+1;
            }
        }
        return -1;
    }
};