[LeetCode] 3Sum

Given an array nums of n integers, are there elements a, b, c in nums 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.

Example:

Given array nums = [-1, 0, 1, 2, -1, -4],

A solution set is:
[
  [-1, 0, 1],
  [-1, -1, 2]
]

三个数字和为0，要求不重复。

首先想到用2Sum的方法，先确定一个数，然后内部循环使用2Sum的哈希算法。最后对结果去重。

结果超时了。贴上代码

class Solution {
public:
    vector<vector<int>> threeSum(vector<int>& nums) {
        vector<vector<int>> res;
        if (nums.empty())
            return res;
        sort(nums.begin(), nums.end());
        for (int i = 0; i < nums.size(); ++i)
        {
            int target = 0-nums[i];
            unordered_set<int> set;
            vector<int> tmp;
            tmp.push_back(nums[i]);
            for (int j = i+1; j < nums.size(); ++j)
            {
                if (set.count(target-nums[j]) > 0)
                {
                    tmp.push_back(target-nums[j]);
                    tmp.push_back(nums[j]);
                }
                if (tmp.size() == 3)
                {
                    res.push_back(tmp);
                    tmp.clear();
                    tmp.push_back(nums[i]);
                    //set.clear();
                }
                set.insert(nums[j]);
            }
        }
        for (auto i = res.begin(); i != res.end(); ++i)
        {
            vector<int> vec1 = *i;
            sort(vec1.begin(), vec1.end());
            for (auto j = i+1; j != res.end(); ++j)
            {
                vector<int> vec2 = *j;
                sort(vec2.begin(), vec2.end());
                if (vec1 == vec2)
                {
                    j = res.erase(j);
                    j--;
                }
            }
        }
        return res;
    }
};

看了看大佬们的答案，豁然开朗

将数组排序，然后确定一个数，再使用二分查找的方法。

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