leetcode 题解：Search in Rotated Sorted Array II （旋转已排序数组查找2）

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.

说明：

1）和1比只是有重复的数字，整体仍采用二分查找

2）方法二 ：

实现：

   一、我的实现：

 class Solution {
 public:
     bool search(int A[], int n, int target) {
         if(n==||n==&&A[]!=target) return false;
         if(A[]==target) return true;
         int i=;
         while(A[i-]<=A[i])  i++;

         bool pre=binary_search(A,,i,target);
         bool pos=binary_search(A,i,n-i,target);
         return pre==false?pos:pre;
     }
 private:
     bool binary_search(int *B,int lo,int len,int goal)
     {
         int low=lo;
         int high=lo+len-;
         while(low<=high)
         {
             int middle=(low+high)/;
             if(goal==B[middle])//找到，返回index
                return true;
             else if(B[middle]<goal)//在右边
                low=middle+;
             else//在左边
                high=middle-;
         }
         return false;//没有，返回-1
     }
 };

二、网上开源实现：

 class Solution {
 public:
     bool search(int A[], int n, int target) {
          int low=;
          int high=n-;
          while(low<=high)
          {
             const int middle=(low+high)/;
             if(A[middle]==target) return true;
             if(A[low]<A[middle])
             {
                if(A[low]<=target&&target<A[middle])
                   high=middle-;
                else
                   low=middle+;
             }
             else if(A[low]>A[middle])
             {
                if(A[middle]<target&&target<=A[high])
                   low=middle+;
                else
                   high=middle-;
             }
             else
                 low++;
          }
          return false;
     }
 };