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3Sum.java
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/*
Author: King, wangjingui@outlook.com
Date: Dec 20, 2014
Problem: 3Sum
Difficulty: Medium
Source: https://oj.leetcode.com/problems/3sum/
Notes:
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:
Elements in a triplet (a,b,c) must be in non-descending order. (ie, a <= b <= c)
The solution set must not contain duplicate triplets.
For example, given array S = {-1 0 1 2 -1 -4},
A solution set is:
(-1, 0, 1)
(-1, -1, 2)
Solution: Simplify '3sum' to '2sum' O(n^2). http://en.wikipedia.org/wiki/3SUM
*/
public class Solution {
public List<List<Integer>> threeSum(int[] num) {
List<List<Integer>> res = new ArrayList<List<Integer>>();
Arrays.sort(num);
int N = num.length;
for (int i = 0; i < N-2 && num[i] <= 0; ++i)
{
if (i > 0 && num[i] == num[i-1])
continue; // avoid duplicates
int twosum = 0 - num[i];
int l = i + 1, r = N - 1;
while (l < r)
{
int sum = num[l] + num[r];
if (sum < twosum) ++l;
else if (sum > twosum) --r;
else {
ArrayList<Integer> tmp = new ArrayList<Integer>();
tmp.add(num[i]); tmp.add(num[l]); tmp.add(num[r]);
res.add(tmp);
++l; --r;
while (l < r && num[l] == num[l-1]) ++l; // avoid duplicates
while (l < r && num[r] == num[r+1]) --r; // avoid duplicates
}
}
}
return res;
}
}