[LeetCode] 4Sum-优快云博客

本文介绍了一种算法用于在数组中查找所有满足特定条件的四元组，即四个数之和等于给定的目标值。算法通过排序数组并使用双指针技巧来高效地找到所有符合条件的四元组，同时确保结果集不包含重复的四元组。

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Question : 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)

思路：

Note : 这个跟3sum也挺像。定义一对指针，指向两头。再定义一对指针，指向中间的两个元素。加起来看看跟target比较一下。决定内部指针怎么移动。

用到了Arrays.sort() 方法，该方法是一种改进的快排，时间复杂度最优情况 O(N*logN)

import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashSet;

public class FourSum {

	public ArrayList fourSum (int[] num, int target) {
			HashSet rs = new HashSet();
			int len = num.length;
			Arrays.sort(num);
			if(len <= 3) return new ArrayList(rs);
			
			for(int i = 0; i < len-3; i++) {
			 for(int k = len-1; k > i+2; k--) {
			     int ab = num[i] + num[k];
			     int c = target-ab;
			     int m = i+1;
			     int n = k-1;
			     while(m < n) {
			         int sum = num[m] + num[n];
			         if(sum == c) {
			             ArrayList elem = new ArrayList();
			             elem.add(num[i]);
			             elem.add(num[m]);
			             elem.add(num[n]);
			             elem.add(num[k]);
			             rs.add(elem);
			             m++;
			             n--;
			         }
			         else if(sum < c) {
			             m++;
			         }
			         else n--;
			     }
			 }
			}
			
			return new ArrayList(rs);

		}
}