Search in Rotated Sorted Array II leetcode java

题目:

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.

 

题解:

     这道题与之前Search in Rotated Sorted Array类似,问题只在于存在dupilcate。那么和之前那道题的解法区别就是,不能通过比较A[mid]和边缘值来确定哪边是有序的,会出现A[mid]与边缘值相等的状态。所以,解决方法就是对于A[mid]==A[low]和A[mid]==A[high]单独处理。

     当中间值与边缘值相等时,让指向边缘值的指针分别往前移动,忽略掉这个相同点,再用之前的方法判断即可。

     这一改变增加了时间复杂度,试想一个数组有同一数字组成{1,1,1,1,1},target=2, 那么这个算法就会将整个数组遍历,时间复杂度由O(logn)降到O(n)

 

 实现代码如下:

 1     public boolean search(int [] A,int target){
 2        if(A==null||A.length==0)
 3          return false;
 4         
 5        int low = 0;
 6        int high = A.length-1;
 7       
 8        while(low <= high){
 9            int mid = (low + high)/2;
10            if(target < A[mid]){
11                if(A[mid]<A[high])//right side is sorted
12                  high = mid - 1;//target must in left side
13                else if(A[mid]==A[high])//cannot tell right is sorted, move pointer high
14                  high--;
15                else//left side is sorted
16                  if(target<A[low])
17                     low = mid + 1;
18                  else 
19                     high = mid - 1;
20            }else if(target > A[mid]){
21                if(A[low]<A[mid])//left side is sorted
22                  low = mid + 1;//target must in right side
23                else if(A[low]==A[mid])//cannot tell left is sorted, move pointer low
24                  low++;
25                else//right side is sorted
26                  if(target>A[high])
27                     high = mid - 1;
28                  else
29                     low = mid + 1;
30            }else
31              return true;
32        }
33        
34        return false;
35 } 

Reference: http://blog.csdn.net/linhuanmars/article/details/20588511

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