【组合&元素和问题】4Sum_4sum问题用python-优快云博客

本文链接：https://blog.youkuaiyun.com/wu_george/article/details/22957425

本文探讨了如何寻找数组中四个元素之和等于目标值的所有唯一四元组，并提出了两种解决方案。一种是深度优先搜索（DFS），但存在时间效率问题；另一种通过排序和双指针技巧实现更高效的求解。

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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)

第一反应：DFS，但是超时了，擦擦

public class Solution {
    ArrayList<ArrayList<Integer>> res = new ArrayList<ArrayList<Integer>>();
    
    public void dfs(int [] num, ArrayList<Integer> l, int rs, int pos){
        if(pos == num.length) return;
        
        if(l.size() == 4){
            res.add(l);
            return ;
        }
        for(int i=pos; i<num.length; i++){
            l.add(num[i]);
            dfs(num, new ArrayList(l), rs - num[i], i+1);
            l.remove(l.size()-1);
        }
    }
    
    public ArrayList<ArrayList<Integer>> fourSum(int[] num, int target) {
        if(num == null) return null;
        Arrays.sort(num);
        ArrayList<Integer> l = new ArrayList<Integer>();
        dfs(num, l, target, 0);
        return res;
    }
}

解法二：将4个和转化成两个和问题（两个指针分两个方向指向数组）

三层for循环解决，O(n^3)

public class Solution {
    ArrayList<ArrayList<Integer>> res = new ArrayList<ArrayList<Integer>>();
    
    public ArrayList<ArrayList<Integer>> fourSum(int[] num, int target) {
        if(num == null) return null;
        Arrays.sort(num);
        int len = num.length;
        
        for(int i=0; i<len; i++){
            if(i!=0 && num[i]==num[i-1]) continue;
            
            for(int j=i+1; j<len; j++){
                if(j!=i+1 && num[j]==num[j-1]) continue;
                
                int p = j+1, q = len - 1;
                while(p < q){
                    if(p!=j+1 && num[p]==num[p-1]){
                        p++;
                        continue;
                    }
                    if(q!=len-1 && num[q] == num[q+1]){
                        q--;
                        continue;
                    }
                    
                    int sum = num[i] + num[j] + num[p] + num[q];
                    if(sum == target){
                        ArrayList<Integer> l = new ArrayList<Integer>();
                        l.add(num[i]);
                        l.add(num[j]);
                        l.add(num[p]);
                        l.add(num[q]);
                        res.add(l);
                        p++;
                    }
                    else if(sum < target) p++;
                    else q--;
                }
            }
        }
        
        return res;
    }
}