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[Problem]
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:
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)
[Analysis]
思路与3Sum一样。事实上NSum问题都可以递归到2Sum然后以O(n^(N-1))复杂度解决，有兴趣还可以进一步用分治法优化(两两解决2Sum问题并merge)。
[Solution]
import java.util.ArrayList; import java.util.Arrays; import java.util.List; public class Solution { public List<List<Integer>> fourSum(int[] num, int target) { List<List<Integer>> solution = new ArrayList<List<Integer>>(); if (num.length < 4) { return solution; } Arrays.sort(num); for (int i = 0; i < num.length - 3; i++) { if (i > 0 && num[i] == num[i - 1]) { continue; // if current value is already tested as the first element, skip; } for (int j = i + 1; j < num.length - 2; j++) { if (j > i + 1 && num[j] == num[j - 1]) { continue; } // Do 2Sum int head = j + 1; int tail = num.length - 1; while (head < tail) { if (head > j + 1 && num[head] == num[head - 1]) { head++; continue; } if (tail < num.length - 1 && num[tail] == num[tail + 1]) { tail--; continue; } int sum = num[i] + num[j] + num[head] + num[tail]; if (sum == target) { solution.add(new ArrayList<Integer>(Arrays.asList(num[i], num[j], num[head], num[tail]))); head++; tail--; } else if (sum > target) { tail--; } else { head++; } } } } return solution; } }
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原文地址：http://www.cnblogs.com/zhangqieyi/p/4865496.html
