【leetcode】Search in Rotated Sorted Array II

Search in Rotated Sorted Array II

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.

 

与I类似，只是在二分搜索前，只是在判断左右两边是否有序时更加麻烦一些。

此时复杂度是O(n)了，和采用线性搜索的方法基本一致

 

 class Solution {
 public:
     bool search(int A[], int n, int target) {

         int left=;
         int right=n-;
         int mid;
         while(left<=right)
         {

             mid=(left+right)/;
             if(A[mid]==target)
             {
                 return true;
             }

             //增加了这些判断语句
             bool isLeft=true;

             for(int i=left;i<=mid;i++)
             {
                 if(A[i]>A[mid])
                 {
                     isLeft=false;
                     break;
                 }
             }

             if(isLeft)
             {
                 if(A[left]<=target&&target<A[mid])
                 {
                     right=mid-;
                 }
                 else
                 {
                     left=mid+;
                 }
             }
             else
             {
                 if(A[mid]<target&&target<=A[right])
                 {
                     left=mid+;
                 }
                 else
                 {
                     right=mid-;
                 }
             }

         }
         return false;

     }
 };