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.

[解题思路]

确实有影响。比如，上一题( )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])  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:       }