leetcode_81_Search in Rotated Sorted Array II

欢迎转载，如有错误或疑问请留言纠正，谢谢

Search in Rotated Sorted Array 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.

//方法：比较笨的class Solution {public:    bool search(vector<int>& nums, int target) {        if(nums.size() == 0)            return false;        if(nums.size() == 1)            if(nums[0] == target)                return true;        int left = 0;        int right = nums.size()-1;        int mid = (right + left) / 2;        bool ans = false;                while(left < right)        {            mid = (right + left) / 2;            while((nums[left] == nums[right]) || (left != mid && nums[mid] == nums[left]))  //left==right不要也是可以的            {                left++;                mid = (right + left) / 2;            }            while(nums[right] == nums[mid])            {                right--;                mid = (right + left) / 2; // return false：[3,1,1], 1            }                            mid = (right + left) / 2;            if(nums[mid] == target)                return true;            else if(right - left == 1)            {                if(nums[right] == target)                    return true;                break;            }            else if(nums[mid] > nums[left])            {                if(nums[mid] > target && nums[left] <= target)                    right = mid;                else                    left = mid;            }            else if(nums[mid] < nums[right])            {                if(nums[mid] < target && nums[right] >= target)                    left = mid;                else                    right = mid;            }        }                return ans;    }};

//方法一：自测Accepted//classified discussion//1. Based on the property of rotated array, there may or may not have one sorted sequence //when one sequence is divided into two parts  //2. make decision under all these casesclass Solution {public:    bool search(int A[], int n, int target) {        if( n==0 )return false;        int mid = 0;int left=0, right=n-1;while( left <= right){mid = left + (right-left)/2 ;if(A[mid] == target)return true;if( A[left] < A[mid] ) //left side sorted, including mid, it is sorted in [l,mid]{if( A[left] <= target && target < A[mid] )right = mid - 1;elseleft = mid + 1;}else if ( A[mid] < A[right] ) //right side sorted, [mid,r] must be sorted, if [l,mid] not sorted{if ( A[mid] < target && target <= A[right] )left = mid + 1;elseright = mid - 1;}else if ( A[left] == A[mid] ) //A[m] is not the target, so remove en element equal to A[m] is safeleft++;else if ( A[right] == A[mid] ) //dittoright--;}return false;    }};