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[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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    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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[LeetCode] Remove Duplicates from Sorted Array II