15. 3Sum

题目要求结果里不能出现重复的结果。

Example 1:

Input: nums = [-1,0,1,2,-1,-4]
Output: [[-1,-1,2],[-1,0,1]]
Explanation: 
nums[0] + nums[1] + nums[2] = (-1) + 0 + 1 = 0.
nums[1] + nums[2] + nums[4] = 0 + 1 + (-1) = 0.
nums[0] + nums[3] + nums[4] = (-1) + 2 + (-1) = 0.
The distinct triplets are [-1,0,1] and [-1,-1,2].
Notice that the order of the output and the order of the triplets does not matter.

Example 2:

Input: nums = [0,1,1]
Output: []
Explanation: The only possible triplet does not sum up to 0.

Example 3:

Input: nums = [0,0,0]
Output: [[0,0,0]]
Explanation: The only possible triplet sums up to 0.

Constraints:

• 3 <= nums.length <= 3000

• -105 <= nums[i] <= 105

My Solutions：

先把nums排序。这样在遍历数组时可以跳过重复数字。对每一个遍历的数字，在ta的后面的数字中用双指针进行搜索。

Time: O(N^2)

Space: O(N)

class Solution {
    public List<List<Integer>> threeSum(int[] nums) {
        List<List<Integer>> res = new ArrayList<>();
        Arrays.sort(nums);
        for (int i = 0; i < nums.length - 2; i++) {
            if (i == 0 || nums[i] != nums[i - 1]) { // 跳过重复数字
                int l = i + 1, r = nums.length - 1, sum = -nums[i];
                while (l < r) {
                    if (nums[l] + nums[r] == sum) {
                        res.add(Arrays.asList(nums[i], nums[l], nums[r]));
                        while (l < r && nums[l] == nums[l + 1]) l++; // skip the duplicate
                        while (l < r && nums[r] == nums[r - 1]) r--; // skip the duplicate
                        l++; // skip one more time to make sure the next number is instinct
                        r--; // skip one more time to make sure the next number is instinct
                    }
                    else if (nums[l] + nums[r] < sum) l++;
                    else r--;
                }
            }
        }
        return res;
    }
}