LeetCode OJ

题目：

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)

解题思路：

　　先将数组排序。 然后分别固定第一个元素为num[0], num[n]。 用2Sum的方法找剩下的两个数。

代码：

public class Solution {
    public ArrayList<ArrayList<Integer>> threeSum(int[] num) {
        Arrays.sort(num);
        ArrayList<ArrayList<Integer>> ans = new ArrayList<ArrayList<Integer>>();
        for (int first = 0; first < num.length;) {
            int left = first + 1, right = num.length - 1;
            while (left < right) {
                int sum = num[first] + num[left] + num[right];
                if (sum == 0) {
                    ArrayList<Integer> cur_ans = new ArrayList<Integer>();
                    cur_ans.add(num[first]); cur_ans.add(num[left]); cur_ans.add(num[right]);
                    ans.add(cur_ans);
                    left++; right--;
                } else if (sum > 0) {
                    right --;
                } else {
                    left ++;
                }
                while (left < right && (left != first+1 && num[left] == num[left-1])) left ++; // delete duplicate triplets
                while (left < right && (right != num.length-1 && num[right] == num[right+1])) right --;// delete duplicate triplets
            }
            for (first++; first < num.length && num[first] == num[first-1]; first ++);// delete duplicate triplets
        }
        return ans;
    }
}