33. Search in Rotated Sorted Array

problem description

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

algorithm thought

其实是二分查考的变种，一个有序数组，rotate之后数组一半还是顺序的。二分查找只要确定一半是否满足 就可以排除另一半。我们找到有序的一半之后，判断target是否期中，如果是就进入这一半查找否则进入另一半。这样每次排除一半，时间复杂度就是O(nlgn)

code

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

algorithm thought

虽然循环中间处理比二分查找多很多步骤，但是时间复杂度还是O(1)的。每次排除一半，最后时间复杂度是O(nlgn)