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Given a sorted array, remove the duplicates in place such that each element appear only once
and return the new length.
Do not allocate extra space for another array, you must do this in place with constant memory.
For example, Given input array A = [1,1,2],
Your function should return length = 2, and A is now [1,2]

//code1:
O(n),O(1)
class Solution{
    public:
        int removeDuplicates(vector<int>& nums){
            if(nums.empty()) return 0;
            
            int index = 0;
            for(int i = 1; i < nums.size(); ++i){
                if(nums[index] != nums[i])
                    nums[++index] = nums[i];
            }
            
            return index + 1;
        }
};


//code2:
class Solution{
    public:
        int removeDuplicates(vector<int>& nums){
            return distance(nums.begin(),unique(nums.begin(),nums.end()));
        }
};

//code3:
class Solution{
    public:
        int removeDuplicates(vector<int>& nums){
            return distance(nums.begin(),removeDuplicates(nums.begin(),nums.end(),nums.begin()));
        }
    
    template<typename InIt, typename OutIt>
    OutIt removeDuplicates(InIt first,InIt last, OutIt output){
        while(first != last){
            *output++ = *first;
            first = upper_bound(first,last,*first);
        }
        return output;
    }
};

 

posted on 2019-09-02 23:00  Worty  阅读(144)  评论(0编辑  收藏  举报