package cn.edu.xidian.sselab.hashtable; import java.util.ArrayList; import java.util.Arrays; import java.util.List; /** * * @author zhiyong wang * title: 4Sum * content: * 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) * */ public class Sum4 { //自己只想到了循环四次,这种方法时间超时,下面是参考大牛的做法 //时间复杂度为O(n^3),第一次循环,选择第一个点;第二次循环,选择第二个点,第三次循环,取出最大值与最小值,然后判断四个值之和是否为target,如果小于,最小值加,反之,最大值减 //学习到了一个新方法Arrays.sort()对数组进行排序,Arrays.asList()新建一个ArrayList对象 public List<List<Integer>> fourSum(int[] nums, int target){ List<List<Integer>> list = new ArrayList<List<Integer>>(); int length = nums.length; if(length < 4 || nums == null) return list; Arrays.sort(nums); if(target < 4 * nums[0] || target > 4 * nums[length-1]) return list; for(int i=0;i<length-3;i++){ if(i>0 && nums[i]== nums[i-1]) continue; for(int j=i+1;j<length-2;j++){ if(j>i+1 && nums[j]==nums[j-1]) continue; int low = j+1; int high = length-1; while(low < high){ int sum = nums[i]+nums[j]+nums[low]+nums[high]; if(sum == target){ list.add(Arrays.asList(nums[i],nums[j],nums[low],nums[high])); while(low<high && nums[low]==nums[low+1]) low++; while(low<high && nums[high]==nums[high-1]) high--; low++; high--; }else if(sum <target){ low++; }else{ high--; } } } } return list; } public static void main(String[] args) { Sum4 s = new Sum4(); int[] nums = {0,0,0,0}; s.fourSum(nums, 0); } } |