[LeetCode]26. Search in Rotated Array II旋转数组查找II

Follow up for "Search in Rotated Sorted Array":
What if duplicates are allowed?

Would this affect the run-time complexity? How and why?

Write a function to determine if a given target is in the array.

解法：同旋转数组查找I，本题也使用二分查找，只是在nums[mid]和nums[right]（或者nums[right]）比较时若二者相等则left++（或者right--）即可。nums[mid]与nums[left]比较的代码：

class Solution {
public:
    int search(vector<int>& nums, int target) {
        int n = nums.size();
        bool res = false; 
        int left = 0, right = n - 1;
        
        while(left <= right)
        {
            int mid = ((left + right) >> 1);
            if(target == nums[mid])
            {
                    res = true;
                    break;
            }
            else if(nums[mid] > nums[left]) //前半部分有序
            {
                if(target >= nums[left] && target < nums[mid])
                    right = mid - 1;
                else
                    left = mid + 1;
            }
            else if(nums[mid] < nums[left]) //后半部分有序
            {
                if(target > nums[mid] && target <= nums[right])
                    left = mid + 1;
                else
                    right = mid - 1;
            }
            else //nums[mid]==nums[left]的情况，包括mid=left和存在重复值两种情况
                left++;
        }
        return res;
    }
};

nums[mid]与nums[right]比较的代码：

class Solution {
public:
    int search(vector<int>& nums, int target) {
        int n = nums.size();
        bool res = false;
        int left = 0, right = n - 1;
        
        while(left <= right)
        {
            int mid = (left + right) >> 1;
            if(target == nums[mid])
            {
                    res = true;
                    break;
            }
            else if(nums[mid] < nums[right]) //后半部分有序
            {
                if(target > nums[mid] && target <= nums[right])
                    left = mid + 1;
                else
                    right = mid - 1;
            }
            else if(nums[mid] > nums[right]) //前半部分有序
            {
                if(target >= nums[left] && target < nums[mid])
                    right = mid - 1;
                else
                    left = mid + 1;
            }
            else //存在重复值的情况
                right--;
        }    
        return res;
    }
};