[Leetcode] search in rotated sorted array ii 搜索旋转有序数组

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.

题意:查询给定目标值是否在数组中,是search in rotated sorted array的扩展。

思路:因为是上一题的扩展,所以很正常的想到,上题的思路,在本题是否合适。结果发现不能强行的套上去。如:1,1,3,1,target=3,若是按照原来的解法,则应该是左边有序,然而不行;若是if条件判断中间等号,为右边有序,也显然不对。参考了Grandyang的博客才知道,只需在判断上加上判断中间值是否等于最右端值,若是,hi向左移动一个,直到不相等。这样就可以继续保持上题的解法了,具体代码如下:

 1 class Solution {
 2 public:
 3     bool search(int A[], int n, int target) 
 4     {
 5         if(n==0)    return false;
 6         int lo=0,hi=n-1;
 7         while(lo<=hi)
 8         {
 9             int mid=(lo+hi)/2;
10             if(A[mid]==target)
11                 return true;
12             else if(A[mid]<A[hi])
13             {
14                 if(A[mid]<target&&A[hi]>=target)
15                     lo=mid+1;
16                 else
17                     hi=mid-1;
18             }
19             else if(A[mid]>A[hi])
20             {
21                 if(A[lo]<=target&&A[mid]>target)
22                     hi=mid-1;
23                 else
24                     lo=mid+1;
25             }
26             else      //重点理解
27                 hi--;
28         }
29         return false;    
30     }
31 };
原文地址:https://www.cnblogs.com/love-yh/p/7141683.html