3Sum

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)

题目的意思是给定一个数组，找出三个数a,b,c，使得a+b+c=0，保证找到的一组三个数中按照从小到大的顺序排列，并且最后的结果中不包括重复的三个数的组合。

思路：知道了找出两个数相加的算法，然后求三个数相加，可以把问题转换为求两个数相加，那么可以固定a，只要找出b+c=-a即可，由于题目规定了不允许包括重复的组合，难点就是要考虑去重, 去除重复的数字时一定要考虑到left和right的范围，不然很容易出错
i+1<=left<=length-2
i+2<=right<=length-1

public class Solution {    public List<List<Integer>> threeSum(int[] num) {     List<List<Integer>> resultList = new ArrayList<List<Integer>>();        // 先将num排序        Arrays.sort(num);        int length = num.length;        for (int i = 0; i < length; i++) {            int left = i + 1;            int right = length - 1;            int target = 0 - num[i];            while (left < right) {                int sum = num[left] + num[right];                if (sum == target) {                    List<Integer> tmp = new ArrayList<Integer>();                    tmp.add(num[i]);                    tmp.add(num[left]);                    tmp.add(num[right]);                    resultList.add(tmp);                    left++;                    right--;                } else if (sum < target)                    left++;                else                    right--;                while (left < length && left>i+1&& num[left - 1] == num[left]) {                    left++;                }                while (right >= i + 1&& right<length-1 && num[right +1] == num[right]) {                    right--;                }            }            while (i < length - 1 && num[i] == num[i + 1]) {                i++;        }        }        return resultList;    }}