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Thursday, January 3, 2013

[LeetCode] Search in Rotated Sorted Array II 解题报告


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.
» Solve this problem

[解题思路]
确实有影响。比如,上一题(http://fisherlei.blogspot.com/2013/01/leetcode-search-in-rotated-sorted-array.html)的程序中默认如果A[m]>=A[l],那么[l,m]为递增序列的假设就不能成立了,比如如下数据
[1,3,1,1,1]
所以,要是想增强该假设,有两个选择
1. 对于每一个递增序列,遍历之,确认。
2. 找到pivot点,然后确定对应序列搜索。


不写代码了。


Update: 3/18/2013. add the implementation
重新想了一下,其实不需要这么复杂。如果A[m]>=A[l]不能确定递增,那就把它拆分成两个条件
1. A[m]>A[l]  递增
2. A[m] ==A[l] 确定不了,那就l++,往下看一步即可。

实现如下
1:       bool search(int A[], int n, int target) {  
2:            int start = 0;  
3:            int end = n-1;  
4:            while(start <= end)  
5:            {  
6:                 int mid = (start+end)/2;  
7:                 if(A[mid] == target) return true;  
8:                 if(A[start] < A[mid])  
9:                 {  
10:                      if(target>=A[start] && target<A[mid])  
11:                           end = mid-1;  
12:                      else   
13:                           start = mid+1;  
14:                 }  
15:                 else if(A[start] > A[mid])  
16:                 {  
17:                      if(target>A[mid] && target<=A[end])  
18:                           start = mid+1;  
19:                      else  
20:                           end = mid-1;  
21:                 }  
22:                 else //skip duplicate one, A[start] == A[mid]  
23:                      start++;  
24:            }  
25:            return false;  
26:       }  








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