4Sum

4Sum

 

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)

这道题和3sum很像，所以我直接在3sum外面套了一层循环，开始以为会超时来着，因为时间复杂度为O(N^2*logn)但还是没超
开始用四重循环，穷举直接超时了，优化了一下也超时，就用的上面的方法

import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;





//class ListNode {
//      public int val;
//      public ListNode next;
//      ListNode(int x) {
//          val = x;
//          next = null;
//      }
//  }

public class Solution {
    public List<List<Integer>> fourSum(int[] num, int target) {
        List<List<Integer>> result = new ArrayList<List<Integer>>();
        Arrays.sort(num);                            //将数组升序排序
        
        for(int q = 0; q < num.length;q++){
            if(q > 0 && num[q] == num[q - 1])
                continue;
            for(int i = q + 1; i < num.length; i++){
                if(i > q + 1 && num[i] == num[i - 1])
                    continue;                            //去重
                int j = i + 1;
                int k = num.length - 1;
                while(j < k){                            //从i + 1 ~ n - 1中找出两个数等于-num[i]
                    if(j > i + 1 && num[j] == num[j - 1])
                    {
                        j++;
                        continue;
                    }
                    if(k < num.length - 1 && num[k] == num[k + 1]){
                        k--;
                        continue;
                    }                                    //去重
                    int temp = num[i] + num[j] + num[k] + num[q];
                    if(temp > target){                        //结果比0大k前移
                        k--;
                        continue;
                    }
                    else if(temp < target){
                        j++;
                        continue;
                    }else{                                //找到一个解
                        List<Integer> element = new ArrayList<Integer>();
                        element.add(num[q]);
                        element.add(num[i]);
                        element.add(num[j]);
                        element.add(num[k]);
                        result.add(element);
                        j++;
                        //break;                            //退出循环
                    }
                    
                }
            }
        }
        
        
               
        return result;
    }
}