【Leetcode】3Sum

题目链接：https://leetcode.com/problems/3sum/

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)

思路：

先将数组排序，然后设立三个浮标，从左右向中间遍历。注意当某个triple满足条件时，为了避免重复，要移动浮标直到指向的元素发生变化。

算法：

public List<List<Integer>> threeSum(int[] nums) {
        List<List<Integer>> lists = new ArrayList<List<Integer>>();
        Arrays.sort(nums);
        for (int i1 = 0; i1 < nums.length - 2; i1++) {
            if (i1 == 0 || nums[i1] > nums[i1 - 1]) { // 避免重复
                int i2 = i1 + 1;
                int i3 = nums.length - 1;
                while (i2 < i3) {
                    if (nums[i2] + nums[i3] + nums[i1] == 0) {
                        List<Integer> list = new ArrayList<Integer>();
                        list.add(nums[i1]);
                        list.add(nums[i2]);
                        list.add(nums[i3]);
                        while (i2 < i3 && nums[i2] == nums[i2 + 1]) {// 避免跟已有结果重复
                            i2++;
                        }
                        while (i2 < i3 && nums[i3] == nums[i3 - 1]) {// 避免跟已有结果重复
                            i3--;
                        }
                        i2++;
                        i3--;
                        lists.add(list);
                    } else if (nums[i2] + nums[i3] + nums[i1] > 0) {
                        i3--;
                    } else if (nums[i2] + nums[i3] + nums[i1] < 0) {
                        i2++;
                    }
                }
            }
        }
        return lists;
    }