Leetcode 18. 4Sum

待我称王封你为后i 2022-06-02 07:48 156阅读 0赞

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: 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:

> \[  
> \[-1, 0, 0, 1\],  
> \[-2, -1, 1, 2\],  
> \[-2, 0, 0, 2\]  
> \]

--------------------

3Sum 和 3Sum Closest 的扩展，同样思路，加强理解。

K Sum 问题的时间复杂度好像为 O(n^(k-1)) ？！

O(n^3)解法

class Solution {  
        List<List<Integer>> result = new ArrayList<List<Integer>>();  
    
        public List<List<Integer>> fourSum(int[] num, int target) {  
            if (num == null || num.length < 4) return result;  
            Arrays.sort(num);  
    
            for (int i = 0; i < num.length-3; i++) {  
                if (i > 0 && num[i] == num[i-1]) continue;  
                for (int j = i+1; j < num.length-2; j++) {  
                    if (j > i+1 && num[j] == num[j-1]) continue;  
                    findTwo(num, j+1, num.length-1, target, num[i], num[j]);  
                }  
            }  
            return result;  
        }  
    
        public void findTwo(int[] num, int begin, int end, int target, int a, int b) { 
            if (begin < 0 || end >= num.length) return; 
            int l = begin, r = end; 
            while (l < r) {  
                if (a+b+num[l]+num[r] < target) {  
                    l++;  
                } else if (a+b+num[l]+num[r] > target) {  
                    r--; 
                } else {  
                    List<Integer> ans = new ArrayList<Integer>();  
                    ans.add(a);  
                    ans.add(b);  
                    ans.add(num[l]);  
                    ans.add(num[r]);  
                    result.add(ans);  
                    l++;  
                    r--; 
                    while (l < r && num[l] == num[l-1]) l++;  
                    while (l < r && num[r] == num[r+1]) r--; 
                }  
            }  
        }  
    }