15. 3Sum：三数之和为0的组合
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: 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]
]
思路：循环法，先给数组排序->[-4,-1,-1,0,1,2],其中i,j,k三数的下标，i先为最左边的一个数，则i<n-3,此时可以根据求两数之和的方法来进行求解；
对一个有序的数组求两数之和为定值，最简单的方法为j,k分别为数组的两端，主键向中间靠拢；
具体代码如下：
import java.util.*;
public class Solution {
public List<List<Integer>> threeSum(int[] nums) {
List<List<Integer>> lists = new ArrayList<>();
if(nums==null||nums.length==0)
return lists;
Arrays.sort(nums);
int n = nums.length;
int i=0;
while(i<=n-3){
int j=i+1;
int k = n-1;
while(j<k){
if(nums[i]+nums[j]+nums[k]==0){
ArrayList<Integer> list = new ArrayList<Integer>();
list.add(nums[i]);
list.add(nums[j]);
list.add(nums[k]);
lists.add(list);
while(j<k&&nums[j]==nums[j+1])
j++;
while(k>j&&nums[k]==nums[k-1])
k--;
j++;
k--;
}else if(nums[i]+nums[j]+nums[k]<0){
j++;
}else{
k--;
}
}
while(i<n-1&&nums[i]==nums[i+1])
i++;
i++;
}
return lists;
}
}