13 Search in Rotated Sorted Array II

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

这道题的核心是用二分法，间或应用一些其他的技巧，可以说是二分法的进阶版。。。

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

13 Search in Rotated Sorted Array II

原文：https://www.cnblogs.com/xiaoyisun06/p/11380290.html