Leetcode.18 4Sum

Leetcode.18 4Sum

Given an array nums of n integers and an integer target, are there elements a, b, c, and d in nums 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.

Example:

Given array nums = [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]
]

Solution

可以由ksum依次递归到2sum

class Solution {
    public List<List<Integer>> fourSum(int[] nums, int target) {
        if(nums.length < 4){
            return new ArrayList<>();
        }
        List<List<Integer>> res = new ArrayList<>(); 
        Arrays.sort(nums);
        int l = 0;
        while(l<nums.length-3){
          //l-1已经实现过所有的方式，所以l不循环
            if(l>0&&nums[l] == nums[l-1]){
                l++;
                continue;
            }
            int m = l+1;
          //3 sum
            while(m<nums.length-2){
                if(m>l+1&&nums[m] == nums[m-1]){
                    m++;
                    continue;
                }
                int sum = target-nums[l]-nums[m];
                int n = m+1;
                int p = nums.length-1;
              //2 sum
                while(n<p){
                    if(nums[n]+nums[p] == sum){
                        res.add(Arrays.asList(nums[l],nums[m],nums[n],nums[p]));
                        while(n<p && nums[n] == nums[n+1]){
                            n++;
                        }
                        while(n<p && nums[p] == nums[p-1]){
                            p--;
                        };
                        n++;
                        p--;
                    }else if(nums[n]+nums[p]>sum){
                        p--;
                    }else{
                        n++;
                    }
                }
                m++;
            }
            l++;
        }
        return res;
    }
}