在上题基础上，如果元素可以重复的话，那么就需要修改getPos，因为有可能直接得到的最小值不是正好在分界的地方。修改在第34行。

1 int findPos(int A[], int left, int right){ 2     if(left > right) 3         return -1; 4     int mid = (left+right)/2; 5     int result; 6     if(A[left] > A[mid]){ 7         result = findPos(A, left, mid-1); 8         if(result == -1) 9             return mid;10         else11             return A[result]
     
       right)23         return -1;24     int mid = (left+right)/2;25     if(A[mid] == target)26         return mid;27     else if(A[mid] < target)28         return bsearch(A, mid+1, right, target);29     else30         return bsearch(A, left, mid-1, target);31 }32     bool search(int A[], int n, int target) {33         int pos = findPos(A, 0, n-1);34         while(pos > 0 && A[pos] == A[pos-1])35             pos--;36         int result = bsearch(A, 0, pos-1, target);37         if(result == -1)38         result = bsearch(A, pos, n-1, target);39         if(result == -1)40             return false;41         return true;42     }
     

 思路二的修改是如果A[m] == A[l]时，要两边都进行判断

1     bool search2(int A[], int n, int target, int l, int r){ 2         if(n == 0) 3             return false; 4         int m; 5         while(l <= r){ 6             m = (l+r)/2; 7             if(A[m] == target) 8                 return true; 9             else if(A[m] > A[l]){10                 if(target < A[m] && target >= A[l])11                     r = m-1;12                 else13                     l = m+1;14             }15             else if(A[m] < A[l]){16                 if(target > A[m] && target <= A[r])17                     l = m+1;18                 else19                     r = m-1;20             }21             else22                 return search2(A, n, target, l, m-1) || search2(A, n, target, m+1, r);23         }24         return false;25     }26     bool search(int A[], int n, int target) {27         return search2(A, n, target, 0, n-1);28     }