【leetcode刷题笔记】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)
题解：参考3Sum，把4Sum问题转换成3Sum，再转换成2Sum，如下图所示：
代码如下：
1 public class Solution { 2 public List<List<Integer>> fourSum(int[] num, int target) { 3 List<List<Integer>> answer = new ArrayList<List<Integer>>(); 4 if(num == null || num.length == 0) 5 return answer; 6 7 Arrays.sort(num); 8 int length = num.length; 9 for(int i = 0;i < length;i++){ 10 if(i!=0 && num[i]== num[i-1] ) 11 continue; 12 for(int j = i+1;j < length;j++ ){ 13 if(j != i+1 && num[j] == num[j-1] ) 14 continue; 15 int start = j + 1; 16 int last = length - 1; 17 while(start < last){ 18 int sum = num[i]+num[j]+num[start]+num[last]; 19 if(sum == target){ 20 ArrayList<Integer> result = new ArrayList<Integer>(); 21 result.add(num[i]); 22 result.add(num[j]); 23 result.add(num[start]); 24 result.add(num[last]); 25 answer.add(result); 26 start++; 27 last--; 28 while(start < last && num[start-1] == num[start]) 29 start++; 30 } 31 else if(sum < target){ 32 start++; 33 } 34 else { 35 last--; 36 } 37 } 38 } 39 } 40 return answer; 41 } 42 }
