LeetCode Search in Rotated Sorted Array II -- 有重复的旋转序列搜索

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.

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

这道题是二分查找Search Insert Position的变体，思路在Search in Rotated Sorted Array中介绍过了，不了解的朋友可以先看看那道题哈。和Search in Rotated Sorted Array唯一的区别是这道题目中元素会有重复的情况出现。不过正是因为这个条件的出现，出现了比较复杂的case，甚至影响到了算法的时间复杂度。原来我们是依靠中间和边缘元素的大小关系，来判断哪一半是不受rotate影响，仍然有序的。而现在因为重复的出现，如果我们遇到中间和边缘相等的情况，我们就丢失了哪边有序的信息，因为哪边都有可能是有序的结果。假设原数组是{1,2,3,3,3,3,3}，那么旋转之后有可能是{3,3,3,3,3,1,2}，或者{3,1,2,3,3,3,3}，这样的我们判断左边缘和中心的时候都是3，如果我们要寻找1或者2，我们并不知道应该跳向哪一半。解决的办法只能是对边缘移动一步，直到边缘和中间不在相等或者相遇，这就导致了会有不能切去一半的可能。所以最坏情况（比如全部都是一个元素，或者只有一个元素不同于其他元素，而他就在最后一个）就会出现每次移动一步，总共是n步，算法的时间复杂度变成O(n)。代码如下：

转载于:https://www.cnblogs.com/zl1991/p/7079377.html

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