81. 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?

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.

此题受到了Search in Rotated Sorted Array以及Find Minimum in Rotated Sorted Array2的启发,两者结合起来做就可以了,代码如下:

 1 public class Solution {
 2     public boolean search(int[] nums, int target) {
 3         if(nums==null||nums.length==0) return false;
 4         int low = 0;
 5         int high = nums.length-1;
 6         while(low<high){
 7             int mid = low +(high-low)/2;
 8             if(nums[mid]==target) return true;
 9             if(nums[mid]<nums[high]){
10                 if(target>nums[mid]&&target<=nums[high]) low = mid+1;
11                 else high = mid-1;
12             }else if(nums[mid]>nums[high]){
13                 if(target>=nums[low]&&target<nums[mid]) high = mid-1;
14                 else low = mid+1;
15             }else{
16                 high--;
17             }
18         }
19         return nums[low]==target?true:false;
20     }
21 }
22 //run time complexity O(longn), the space complexity could be O(1);

 

这道题开始我想用先找最小元素然后在找制定元素来做,但是没有通过。原因是这时候realmid和realend是相等的,而此时的nums【realmid】是不等于target的,代码如下:

public class Solution {

    public boolean search(int[] nums, int target) {

        if(nums.length==0) return false;

        int left = 0;

        int right = nums.length-1;

        while(left<right){

            int mid = left+(right-left)/2;

            if(nums[mid]<nums[right]){

                right = mid;

            }else if(nums[mid]>nums[right]){

                left = mid+1;

            }else{

                right--;

            }

        }

        int min = left;

        left = 0;

        right = nums.length-1;

        while(left<=right){

            int mid = left+(right-left)/2;

            int realmid = (min+mid)%nums.length;

            if(nums[realmid]==target) return true;

            else if(nums[realmid]>target) right = mid-1;

            else left = mid+1;

        }

        return false;

    }

}

红线部分想改进但是不知道怎么改。

posted @ 2017-02-01 09:09  CodesKiller  阅读(144)  评论(0编辑  收藏  举报