【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;
}