15. 3Sum

题目

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]
]

解析

• 想写出一次能AC的代码真不容易！
• 很多细节问题，和sumtwo不一样的是：这次有重复元素；sumtwo假定没有重复元素，且只需要返回下标值
• 要跳过重复的元素

class Solution_15 {
public:

    void twosum(vector<vector<int>>& vecs,vector<int>& nums,int start, int target)
    {
        vector<int> ans;
        int end = nums.size() - 1;
        while (start<end)
        {
            if (nums[start]+nums[end]==target)
            {
                ans.push_back(-target);
                ans.push_back(nums[start]);
                ans.push_back(nums[end]);
                vecs.push_back(ans);
                ans.clear();
                //start++; end--;  跳不过重复的元素
                while (start<end&&nums[start]==nums[start+1])
                {
                    start++;
                }
                while (start<end&&nums[end]==nums[end-1])
                {
                    end--;
                }
                start++; end--;
            }else if (nums[start] + nums[end] < target)
            {
                start++;
            }
            else
            {
                end--;
            }
        }
        return;
    }

    vector<vector<int>> threeSum(vector<int>& nums) { // Time Limit Exceeded
        vector<vector<int>> vecs;
        if (nums.size() <= 2)
        {
            return vecs;
        }
        sort(nums.begin(), nums.end());
        for (int i = 0; i < nums.size() - 2;++i)
        {
            if (i>0&&nums[i]==nums[i-1]) //忽略掉有重复元素的值
            {
                continue;
            }
            twosum(vecs, nums,i+1, -nums[i]);
        }
        return vecs; //
    }

    vector<vector<int>> threeSum1(vector<int>& nums) { // Time Limit Exceeded
        vector<vector<int>> vecs;
        if (nums.size()<=2)
        {
            return vecs;
        }

        unordered_map<int, int> mp;
        vector<int> ans;
        for (int i = 0; i < nums.size() - 2;++i)
        {
            mp.clear();
            for (int j = i + 1; j < nums.size();++j)
            {
                auto iter = mp.find(-(nums[i] + nums[j])); //查找主关键字
                if (iter!=mp.end())
                {
                    ans.push_back(nums[i]);
                    ans.push_back(iter->first);
                    ans.push_back(nums[j]);
                    sort(ans.begin(), ans.end());  //处理有重复元素
                    if (find(vecs.begin(),vecs.end(),ans)==vecs.end()) //没有元素才插入操作
                    {
                        vecs.push_back(ans);
                    }
                    ans.clear();
                }
                else
                {
                    mp.insert(make_pair(nums[j], j));
                }
            }
        }
        return vecs; //
    }

};

题目来源

• 15. 3Sum