15. 3Sum

https://leetcode.com/problems/3sum/description/

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: 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]

]

Thoughts

找数组中所有和为0的三元组，结果中不能有任何重复。排序后nums[i]作为最大元素的候选，在i元素前找和为-nums[i]的二元对，也就是2sum。2sum用首尾双指针。为了去除结果中重复的，2sum内判断当前指针后面是否为相同元素，是就说明当前值已经考虑过，跳。最外层的i同理，不过判断条件不能再是看前面i -1是否和nums[i]相同，因为这样会错误的把两个nums[i]作为解元素的情况刨除。

Code

/*
 * @lc app=leetcode id=15 lang=cpp
 *
 * [15] 3Sum
 */

// @lc code=start
class Solution {
public:
    vector<vector<int>> threeSum(vector<int>& nums) {
        const int N = nums.size();
        sort(nums.begin(), nums.end());
        vector<vector<int>> res;
        for (int i = 2; i < N; ++i) {
            if (i < N - 1 && nums[i] == nums[i + 1]) continue;
            const int t = -nums[i];
            for (int j = 0, k = i - 1; j < k;) {
                if (nums[j] + nums[k] < t || j > 0 && nums[j] == nums[j - 1]) {
                    ++j;
                } else if (nums[j] + nums[k] > t || k < i - 1 && nums[k] == nums[k + 1]) {
                    --k;
                } else res.push_back({nums[i], nums[j++], nums[k--]});
            }
        } 
        return res;
    }
};
// @lc code=end

class Solution {
    public List<List<Integer>> threeSum(int[] nums) {
        Arrays.sort(nums);
        List<List<Integer>> res = new ArrayList<>();
        for (int i = 0; i < nums.length; i++) {
            if (i > 0 && nums[i] == nums[i - 1]) {
                continue;
            }
            int target = -nums[i];
            int l = i + 1, r = nums.length - 1;
            while (l < r) {
                if (nums[l] + nums[r] < target) {
                    l++;
                } else if (nums[l] + nums[r] > target) {
                    r--;
                } else {
                    res.add(Arrays.asList(nums[i], nums[l], nums[r]));
                    l++;
                    r--;
                    while (l < r && nums[l] == nums[l - 1]) {
                        l++;
                    }
                    while (l < r && nums[r] == nums[r + 1]) {
                        r--;
                    }
                }
            }
        }

        return res;
    }
}

Analysis

Errors

1. 没想到可以直接跳过Nums[i] == nums[i-1]的

2. l和r也得这么跳以防止重复

O(N^2).