15.3Sum

Tags: Array Medium

Links: https://leetcode.com/problems/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]
]

---

Answer:

class Solution {
public:
    vector<vector<int>> threeSum(vector<int>& nums) {
        vector<vector<int>> result;
        int target = 0;

        if (nums.size() < 3)
            return result;

        sort(nums.begin(), nums.end());

        for (int i = 0; i < nums.size() - 2; ++i){
            if (i > 0 && nums[i] == nums[i - 1]) continue;

            int j = i + 1;
            int k = nums.size() - 1;

            while (j < k){
                if (nums[i] + nums[j] + nums[k] < target) ++j;
                else if (nums[i] + nums[j] + nums[k] > target) --k;
                else{
                    result.push_back({nums[i], nums[j], nums[k]});
                    ++j;
                    --k;
                }
            }
        }

        sort(result.begin(), result.end());
        result.erase(unique(result.begin(), result.end()), result.end());

        return result;
    }
};

/*
Runtime: 112 ms, faster than 41.77% of C++ online submissions for 3Sum.
Memory Usage: 16.9 MB, less than 28.24% of C++ online submissions for 3Sum.
*/

class Solution {
public:
    vector<vector<int>> threeSum(vector<int>& nums) {
        vector<vector<int>> result;
        if (nums.size() < 3) return result;
        sort(nums.begin(),nums.end());
        const int target = 0;
        auto last=nums.end();
        for (auto a=nums.begin(); a<prev(last,2); a++) {
            auto b=next(a);//next 1 postion
            auto c=prev(last);//prev 1 postion
            if (a>nums.begin() &&  *a == *prev(a)) continue; //a have processed
            if (*a>0) return result;
            while(b<c) {
                if (*a + *b + *c < target) b++;
                else if (*a + *b + *c > target) c--;
                else {          
                    result.push_back({*a,*b,*c});
                    while (b<c && *b==*(next(b))) b++;//when break,b != next(b),but b==prev(b).
                    while (b<c && *c==*(prev(c))) c--;
                    b++; 
                    c--;   
                }                     
            } 
        }    

        return result;
    }
};
/*
Runtime: 100 ms, faster than 69.55% of C++ online submissions for 3Sum.
Memory Usage: 14.7 MB, less than 91.76% of C++ online submissions for 3Sum.
*/

方法二相较于方法一少了用unique的环节，第一种方法对于中间重复的序列通过sort和erase(unique,c.end())来实现消除，但是会在sort和unique上浪费时间。方法二则通过19，20行来消除重复序列。

这里值得注意的是序列长度小于3则返回空vector，对于一个特殊的序列如[0,0,0,0,0,0,0,...]此时如果没有if (i > 0 && nums[i] == nums[i - 1])会出现超时情况。

class Solution {
public:
    vector<vector<int>> threeSum(vector<int>& nums) {
        std::ios_base::sync_with_stdio(false);
        cin.tie(NULL);
        cout.tie(NULL);

        int n = nums.size();
        vector<vector<int>> res;
        if (n < 3) return res;

        sort(nums.begin(), nums.end());
        for (int i = 0; i < n - 2; ++i) {
            if (i > 0 && nums[i] == nums[i - 1]) continue;

            if (nums[i] > 0) return res;
            int j = i + 1;
            int k = n - 1;
            while (j < k) {
                int sum = nums[i] + nums[j] + nums[k];
                if (sum < 0) ++j;
                else if (sum > 0) --k;
                else {
                    res.push_back({nums[i], nums[j], nums[k]});
                    while (j < k && nums[j] == nums[j + 1]) ++j;
                    while (j < k && nums[k] == nums[k - 1]) --k;
                    ++j; --k;
                }
            }

        }
        return res;
    }
};

Runtime: 88 ms, faster than 98.71% of C++ online submissions for 3Sum.
Memory Usage: 14.1 MB, less than 100.00% of C++ online submissions for 3Sum.