package com.circle.algorithm;
import java.util.*;
/**
* Created by keweiyang on 2017/3/1.
* 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:
* 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)
*/
public class Sum {
public void sum(int[] as, int target) {
List<List<Integer>> list = new ArrayList<>();
for (int i = 0; i < as.length; i++) {
for (int j = i + 1; j < as.length; j++) {
int k = as.length - 1;
while (target != as[i] + as[j] + as[k]) {
if (target < as[i] + as[j] + as[k]) {
k--;
if (j > k) {
break;
}
} else if (target > as[i] + as[j] + as[k]) {
break;
}
}
if (target == as[i] + as[j] + as[k]) {
Integer[] arr = new Integer[3];
arr[0] = as[i];
arr[1] = as[j];
arr[2] = as[k];
list.add(Arrays.asList(arr));
}
}
}
Iterator it = list.iterator();
while (it.hasNext()) {
System.out.println(it.next());
}
}
public static void main(String[] args) {
Sum sum = new Sum();
int[] as = new int[]{-1, 0, 1, 2, -1, -4};
Arrays.sort(as);
sum.sum(as, 0);
}
}