Search in Rotated Sorted Array II leetcode java

本文轉載自查看原文 2014-07-22 00:40 3075 leetcode每題整理

題目：

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.

題解：

這道題與之前Search in Rotated Sorted Array類似，問題只在於存在dupilcate。那么和之前那道題的解法區別就是，不能通過比較A[mid]和邊緣值來確定哪邊是有序的，會出現A[mid]與邊緣值相等的狀態。所以，解決方法就是對於A[mid]==A[low]和A[mid]==A[high]單獨處理。

當中間值與邊緣值相等時，讓指向邊緣值的指針分別往前移動，忽略掉這個相同點，再用之前的方法判斷即可。

這一改變增加了時間復雜度，試想一個數組有同一數字組成{1，1，1，1，1}，target=2, 那么這個算法就會將整個數組遍歷，時間復雜度由O(logn)升到O(n)

實現代碼如下：

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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