[LeetCode] Remove Duplicates from Sorted Array II

Follow up for "Remove Duplicates":
What if duplicates are allowed at most twice?

For example,
Given sorted array A = [1,1,1,2,2,3],

Your function should return length = 5, and A is now [1,1,2,2,3].

 

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  Array Two Pointers

 

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    这个很容易判断了，设两个index，一个是遍历的，一个是指向返回的最后位置，因为已经排序了，所以判断索引的值与最后位置及前一个的值是否相等，相等继续遍历，不相等更新索引。

 1 #include <iostream>

 2 using namespace std;

 3 

 4 class Solution {

 5 public:

 6     int removeDuplicates(int A[], int n) {

 7         if(n<3) return n;

 8 //        cout<<n<<endl;

 9         int retidx= 1,curidx =2;

10         for(;curidx<n;curidx++){

11 //            cout<<retidx<<" "<<curidx<<endl;

12             if(A[retidx]==A[curidx]&&A[retidx-1]==A[curidx])

13                 continue;

14             A[++retidx] = A[curidx];

15         }

16         return retidx+1;

17     }

18 };

19 

20 int main()

21 {

22     int a[]={};

23     Solution sol;

24     int ret = sol.removeDuplicates(a,sizeof(a)/sizeof(int));

25     for(int i=0;i<ret;i++)

26         cout<<a[i]<<" ";

27     cout<<endl;

28     return 0;

29 }

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