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[Problem]
Given an array S of n integers, are there elements a, b, c in S such that a + b + c = 0? Find all unique triplets in the array which gives the sum of zero.
Note:
For example, given array S = {-1 0 1 2 -1 -4},
A solution set is:
(-1, 0, 1)
(-1, -1, 2)
[Analysis]
这题的要点是Note里面的两个要求，这个可以通过排序解决。排序之后先提出一个数字再用2Sum解决即可。需要注意的是跳过已经用过的数字。总体复杂度是O(nlogn) + O(n*n) = O(n*n)。
[Solution]
import java.util.ArrayList; import java.util.Arrays; import java.util.List; public class Solution { public List<List<Integer>> threeSum(int[] num) { List<List<Integer>> solution = new ArrayList<List<Integer>>(); if (num.length < 3) { return solution; } Arrays.sort(num); int target = 0; for (int i = 0; i < num.length - 2; i++) { if (i > 0 && num[i] == num[i - 1]) { continue; // if current value is already tested as the first element, skip; } // Do 2Sum int head = i + 1; int tail = num.length - 1; while (head < tail) { if (head > i + 1 && num[head] == num[head - 1]) { head++; continue; // if current value is already tested as the second, skip; } if (tail < num.length - 1 && num[tail] == num[tail + 1]) { tail--; continue; // similar test for the third element } int sum = num[i] + num[head] + num[tail]; if (sum == target) { solution.add(new ArrayList<Integer>(Arrays.asList(num[i], 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/4865423.html
