Leetcode刷题记——18. 4Sum(4个数字和)

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一、题目叙述：

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]]

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二、解题思路：

搞不懂，我用了和15题一模一样的算法，就把15题3Sum的算法加了个循环和防止重复的判断，就过了。。。。。。

（1）主体思路：循环前三个数，然后二分查找（所以数组要先排序）为target减去前三数和的第四个数，这样复杂度能从O（n4）减到O（n3logn）。

完全照搬15题。。。

三、源码：

import java.util.ArrayList;  import java.util.Arrays;  import java.util.List;    public class Solution   {      List<List<Integer>> ans = new ArrayList<List<Integer>>();      public List<List<Integer>> fourSum(int[] nums, int target)       {          Arrays.sort(nums);          int q;          int h;          //int count = 0;          //List<List<Integer>> reslist = new ArrayList<List<Integer>>();          for (int i = 0; i < nums.length; i ++)          {              if (i > 0 && nums[i] == nums[i-1]) continue;//去除重复              for (int j = i + 1; j < nums.length; j++)              {                  if (j > i+1 && nums[j] == nums[j-1]) continue;//去除重复                  for (int k = j + 1; k < nums.length; k++)                {                if (k > j+1 && nums[k] == nums[k-1]) continue;                 q = BinarySearch(target-nums[i]-nums[j]-nums[k], nums, k+1, nums.length - 1);                       if (q > k)                       {                           //count ++;                           List<Integer> list = new ArrayList<Integer>();                           list.add(nums[i]);                           list.add(nums[j]);                           list.add(nums[k]);                         list.add(nums[q]);                           ans.add(list);                       }                  }                           }          }                    return ans;      }      public int BinarySearch (int k,int[] num,int first, int last)      {          //int first = 0;          //int last = num.length - 1;          int mid;          while (first <= last)          {              mid = (first + last) / 2;              if (num[mid] > k) last =mid - 1;              else if (num[mid] < k) first = mid + 1;              else return mid;          }          return -1;      }      public static void main(String args[])      {          int[] a = {1, 0, -1, 0, -2, 2};          Solution so = new Solution();          System.out.println(so.fourSum(a, 0));                                    }  }

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