LeetCode 18. 4Sum

原题链接在这里：https://leetcode.com/problems/4sum/

题目：

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]
]

题解：

这道题其实是3Sum的扩展，使用的方法基本相同，只是多加了一层loop.

但要注意一点：For inner j loop, if statement, 判断overflow 时要写成 j>i+1, 而不是j>0, 与 j 的 初始条件有关。若是写成j>0的话，含有四个相同数的输入就会被跳过. e.g. 0,0,0,0 target = 0.

Time Complexity: O(n^3), 两层for, 每一层用时n, 内层for里面的while 用了O(n) time, 一共用了O(n^3).

Space: O(1), regardless res.

AC Java:

class Solution {
    public List<List<Integer>> fourSum(int[] nums, int target) {
        List<List<Integer>> res = new ArrayList<>();
        if(nums == null || nums.length < 4){
            return res;
        }
        
        Arrays.sort(nums);
        int n = nums.length;
        for(int i = 0; i < n - 3; i++){
            if(i > 0 && nums[i] == nums[i - 1]){
                continue;
            }
            
            for(int j = i + 1; j < n - 2; j++){
                if(j > i + 1 && nums[j] == nums[j - 1]){
                    continue;
                }
                
                int p = j + 1;
                int q = n - 1;
                while(p < q){
                    long sum = (long)nums[i] + (long)nums[j] + (long)nums[p] + (long)nums[q];
                    if(sum < (long)target){
                        p++;
                    }else if(sum > (long)target){
                        q--;
                    }else{
                        res.add(Arrays.asList(nums[i], nums[j], nums[p], nums[q]));
                        p++;
                        q--;
                        
                        while(p < q && nums[p] == nums[p - 1]){
                            p++;
                        }
                        
                        while(p < q && nums[q] == nums[q + 1]){
                            q--;
                        }
                    }
                }
            }
        }
        
        return res;
    }
}

类似4Sum II.