Search in rotated array two

description:

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

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

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).

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

The array may contain duplicates.

Thoughts:

1.按照写rotated sorteg array时的思路，我写这个问题的时候，在重新设置low和high的时候，跳过了和middle一样的重复部分。另外要注意的一个点就是当nums[low]和nums[high]相等的时候我们要重新设置low和high

public boolean search(int[] nums, int target){
        int low = 0;
        int high = nums.length-1;
        if(nums[high] == nums[low]&&nums.length>2){
            for(int i = high-1;i>=low;i--){
                if(nums[i] == nums[high]){
                    high--;
                }
            }
        }
        while(low <= high){
            int middle = (low+high)/2;
            if(nums[middle] == target){
                return true;
            }else if(nums[middle] >=nums[low]){
                if(target>=nums[low] && target<nums[middle]){
                    for(int i = middle-1;i>=low;i--){
                        if(nums[i] == nums[middle]){
                            middle--;
                        }else{
                            break;
                        }
                    }
                    high = middle -1; 
                }else{
                    for(int i = middle+1;i<=high;i++){
                        if(nums[i] == nums[middle]){
                            middle++;
                        }else{
                            break;
                        }
                    }
                    low = middle + 1;
                }
            }else{
                if(target>nums[middle]&&target<=nums[high]){
                    for(int i =middle+1;i<=high;i++){
                        if(nums[i] == nums[middle]){
                            middle++;
                        }else{
                            break;
                        }
                    
                    }
                    low = middle +1;
                }else{
                    for(int i = middle-1;i>=low;i--){
                        if(nums[i] == nums[middle]){
                            middle--;
                        }else{
                            break;
                        }
                        
                    }
                    high = middle - 1;
                }
            }
        }
        return false;
    }

2.前面的解法，思路是清晰的过程是麻烦的，需要加很多的判断条件，为了避免这个问题，我们不讲middle和low进行比较，而是让它和high进行比较，这样就能够避免之前的nums[low]和nums[high]相等的情况，所带来的麻烦。

public boolean search2(int[] nums, int target){
        if(nums.length == 0){
            return false;
        }
        int low = 0;
        int high = nums.length - 1;
        while(low < high){
            int middle = (low +high)/2;
            if(nums[middle] == target){
                return true;
            }else{
                if(nums[middle] < nums[high]){
                    if(target > nums[middle] && target <= nums[high]){
                        low = middle + 1;
                    }else{
                        high = middle - 1;
                    }
                }else if(nums[middle] > nums[high]){
                    if(target >= nums[low] & target < nums[middle]){
                        high = middle -1;
                    }else{
                        low = middle +1;
                    }
                }else{
                    high--;
                }
            }
        }
        return nums[low]==target;
    }