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  • 3Sum

    题目描述:

    Given an array S of n integers, are there elements abc in S such that a + b + c = 0? Find all unique triplets in the array which gives the sum of zero.

    Note:

    • Elements in a triplet (a,b,c) must be in non-descending order. (ie, a ≤ b ≤ c)
    • The solution set must not contain duplicate triplets.
        For example, given array S = {-1 0 1 2 -1 -4},
    
        A solution set is:
        (-1, 0, 1)
        (-1, -1, 2)

      读完题目,立刻想到之前做过的Two Sum,原理相同。只是我提交的时候遇到OLE(Output Limit Exceeded),原因是输出的结果中包含重复的 triplet。

    solution:

    vector<vector<int> > threeSum(vector<int> &num) {
        int n = num.size();
        vector<vector<int> > res;
        if(n < 3)
            return res;
        sort(num.begin(),num.end());
        vector<int> temp(3,0);
        int low, high;
        for (int i = 0;i < n-2;++i)
        {
            low = i + 1;
            high = n - 1; 
            while (low < high)
            {
                if(num[i] + num[low] + num[high] == 0)
                {
                    temp[0] = num[i];
                    temp[1] = num[low];
                    temp[2] = num[high];
                    res.push_back(temp);
                    ++low;
                    --high;
                    while (low < high && num[low] == num[low-1])
                        ++low;
                    while (low < high && num[high] == num[high+1])
                        --high;
                }
                else if (num[i] + num[low] + num[high] < 0)
                    ++low;
                else
                    --high;
            }
            while (i+1 < n-2 && num[i+1] == num[i])
                ++i;
        }
        return res;
    }

      想要AC,一是思路,二是细节。Fighting!!!

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  • 原文地址:https://www.cnblogs.com/gattaca/p/4281581.html
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