Saturday, November 15, 2014

[LeetCode] 4Sum

Given an array S of n integers, are there elements abc, 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)



思路:

既然已经做了2sum, 3sum, 3sum closest。自然能推广到4sum。但这里希望能推广到更普遍的k-sum问题。这里使用递归的思路:

1. k-sum问题可以转化为(k-1)-sum问题:对于数组中每个数A[i],在A[i+1:n-1]中寻找target-A[i]的(k-1)-sum问题。
2. 直到k=2时,用2sum的双指针扫描来完成。

去重复解的技巧和3Sum问题一模一样。


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class Solution {
public:
    vector<vector<int> > fourSum(vector<int> &num, int target) {
        vector<vector<int>> allSol;
        vector<int> sol;
        sort(num.begin(),num.end());
        kSum(num, 0, num.size()-1, target, 4, sol, allSol);
        return allSol;
    }
    
    void kSum(vector<int> &num, int start, int end, int target, int k, vector<int> &sol, vector<vector<int>> &allSol) {
        if(k<=0) return;
        if(k==1) {
            for(int i=start; i<=end; i++) {
                if(num[i]==target) {
                    sol.push_back(target);
                    allSol.push_back(sol);
                    sol.pop_back();
                    return;
                }
            }
        } 
        
        if(k==2) {
            twoSum(num, start, end, target, sol, allSol);
            return;
        }
    
        for(int i=start; i<=end-k+1; i++) {
            if(i>start && num[i]==num[i-1]) continue;
            sol.push_back(num[i]);
            kSum(num, i+1, end, target-num[i], k-1, sol, allSol);
            sol.pop_back();
        }
    }
    
    void twoSum(vector<int> &num, int start, int end, int target, vector<int> &sol, vector<vector<int>> &allSol) {
        while(start<end) {
            int sum = num[start]+num[end];
            if(sum==target) {
                sol.push_back(num[start]);
                sol.push_back(num[end]);
                allSol.push_back(sol);
                sol.pop_back();
                sol.pop_back();
                start++;
                end--;
                while(num[start]==num[start-1]) start++;
                while(num[end]==num[end+1]) end--;
            }
            else if(sum<target)
                start++;
            else
                end--;
        }
    }
};

1 comment:

  1. 那用这个方法的时间复杂度是多少呢?

    ReplyDelete