【JAVA、C++】LeetCode 015 3Sum

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：

3Sum对初学者估计是丈二和尚摸不着头脑，其实NSum都有固定的解法。参考链接：

http://tech-wonderland.net/blog/summary-of-ksum-problems.html

基本上所有思路都是排序后查找，防止出现List的重复，因此想到使用set类型，JAVA实现如下：

static public List<List<Integer>> threeSum(int[] num) {
		LinkedHashSet<List<Integer>> set = new LinkedHashSet<List<Integer>>();
		if (num.length <= 2)
			return new ArrayList<List<Integer>>(set);
		final int tar = 0;
		Arrays.sort(num);
		for (int i = 0; i < num.length - 2; i++) {
			int j = i + 1, k = num.length - 1;
			while (j < k) {
				if (num[i] + num[j] + num[k] < tar)
					j++;
				else if (num[i] + num[j] + num[k] > tar)
					k--;
				else {
					set.add(Arrays.asList(num[i], num[j], num[k]));
					j++;
					k--;
				}
			}
		}
		return new ArrayList<List<Integer>>(set);
	}

 结果第一次提交Time Limit Exceeded，第二次提交竟然神奇般的通过了！！！时间425ms，估计也是擦线飘过。

 原因在于set每添加一个元素都会和set中元素进行比较，也就是说需要一次遍历，这样每添加一个元素，时间开销就会比较大，因此修改代码如下，时间322 ms：

static public List<List<Integer>> threeSum(int[] num) {
    ArrayList<List<Integer>> list = new ArrayList<List<Integer>>();
		final int tar = 0;
		Arrays.sort(num);
		for (int i = 0; i < num.length - 2; i++) {
			int j = i + 1, k = num.length - 1;
			while(i < k && num[i]==num[i+1])
				i++;
			while (j < k) {
				if (num[i] + num[j] + num[k] < tar)
					j++;
				else if (num[i] + num[j] + num[k] > tar)
					k--;
				else {
					list.add(Arrays.asList(num[i], num[j], num[k]));
					j++;
					k--;
					while(j<k&&num[j]==num[j-1])
						j++;
					while(k>j&&num[k]==num[k+1])
						k--;
				}
			}
		}
		return list;
	}

 C++：

class Solution {
public:
    vector<vector<int>> threeSum(vector<int>& nums) {
        vector<vector<int>> res;
        if (nums.size() <= 2)
            return res;
        const int tar = 0;
        sort(nums.begin(),nums.end());
        for (int i = 0; i < nums.size() - 2; i++) {
            int j = i + 1, k = nums.size() - 1;
            while (i < k && nums[i] == nums[i + 1])
                i++;
            while (j < k) {
                if (nums[i] + nums[j] + nums[k] < tar)
                    j++;
                else if (nums[i] + nums[j] + nums[k] > tar)
                    k--;
                else {
                    res.push_back({ nums[i], nums[j], nums[k]});
                    j++;
                    k--;
                    while (j<k&&nums[j] == nums[j - 1])
                        j++;
                    while (k>j&&nums[k] == nums[k + 1])
                        k--;
                }
            }
        }
        return res;
    }
};