【LeetCode】4Sum

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4Sum

Given an array S of n integers, are there elements a, b, c, and d in S such that a + b + c + d = target? Find all unique quadruplets in the array which gives the sum of target.

Note:

• Elements in a quadruplet (a,b,c,d) must be in non-descending order. (ie, a ≤ b ≤ c ≤ d)
• The solution set must not contain duplicate quadruplets.

 

    For example, given array S = {1 0 -1 0 -2 2}, and target = 0.    A solution set is:    (-1,  0, 0, 1)    (-2, -1, 1, 2)    (-2,  0, 0, 2)

 

先確定前兩個數num[i],num[j]，

然後設定雙指標k,l分別指向兩端，往中間掃。

(1)(sum = num[i]+num[j]+num[k]+num[l]) == taget，則找到其中一個解（注意用map去重）。k++,l--.

(2)sum > target, l--

(3)sum < target, k++

 

class Solution {public:    vector<vector<int> > fourSum(vector<int> &num, int target) {        vector<vector<int> > result;        if(num.empty() || num.size() < 4)            return result;        int size = num.size();        sort(num.begin(), num.end());        map<vector<int>, bool> m;        for(int i = 0; i < size-3; i ++)        {            for(int j = i+1; j < size-2; j ++)            {                int k = j+1;    //k < size-1                int l = size-1;                while(k < l)                {                    int sum = num[i]+num[j]+num[k]+num[l];                    if(sum == target)                    {                        vector<int> cur(4,0);                        cur[0] = num[i];                        cur[1] = num[j];                        cur[2] = num[k];                        cur[3] = num[l];                        if(m.find(cur) == m.end())                        {                            result.push_back(cur);                            m[cur] = true;                        }                        k ++;                        l --;                    }                    else if(sum > target)                        l --;                    else                        k ++;                }            }        }        return result;    }};

【LeetCode】4Sum