15.Three Sum

思路

• 直接暴力，(O(n^4 log(n)))，不出意外的超时了。。

class Solution {
public:
    vector<vector<int>> threeSum(vector<int>& nums) {
        vector<vector<int>> res;
        
        int i,j,k;
        int len = nums.size();
        for(i = 0; i < len; i++){
            for(j = i+1; j < len; j++){
                for(k = j+1; k < len; k++){
                    if(nums[i]+nums[j]+nums[k] == 0){
                        vector<int> inRes;
                        inRes.push_back(nums[i]);
                        inRes.push_back(nums[j]);
                        inRes.push_back(nums[k]);
                        bool ok = true;
                        for(int m = 0; m < res.size(); m++){
                            if(issame(inRes,res[m])){ 
                                ok = false;
                                continue;
                            }
                        }
                        if(ok)  res.push_back(inRes);
                    }
                }
            }
        }
        return res;
    }
    bool issame(vector<int> a,vector<int> b){
        if(a.size() != b.size()) return false;
        sort(begin(a),end(a));
        sort(begin(b),end(b));
        bool ok = true;
        for(int i = 0; i < a.size(); i++){
            if(a[i] != b[i]){
                ok = false;
            }
        }
        return ok;
    }
};

• 利用map暴力，时间复杂度 (O(n^3 log(n)))，依然超时。。

class Solution {
public:
    vector<vector<int>> threeSum(vector<int>& nums) {
        vector<vector<int>> res;
        map<int,int> tmp;
        int i,j,k;
        int len = nums.size();
        for(i = 0; i < len; i++){
            tmp[nums[i]] = i;
        }
        for(i = 0; i < len; i++){
            for(j = i+1; j < len; j++){
                    vector<int> inRes = {};
                    if(tmp.find(-(nums[i]+nums[j])) != tmp.end() && tmp[-(nums[i]+nums[j])] != i && tmp[-(nums[i]+nums[j])] != j){
                            inRes.push_back(nums[i]);
                            inRes.push_back(nums[j]);
                            inRes.push_back(-(nums[i]+nums[j]));  
                            bool ok = true;
                        for(int m = 0; m < res.size(); m++){
                            if(issame(inRes,res[m])) ok = false;   
                        }
                        if(ok){
                            res.push_back(inRes);
                         }
                    }
                    
        }
    }
    return res;
}
    bool issame(vector<int> a,vector<int> b){
        if(a.size() != b.size()) return false;
        sort(begin(a),end(a));
        sort(begin(b),end(b));
        bool ok = true;
        for(int i = 0; i < a.size(); i++){
            if(a[i] != b[i]){
                ok = false;
            }
        }
        return ok;
    }
};

• 利用3个指针，首先固定一个，然后类似于Two Sum问题。(这里是用map存储返回结果的，需要避免重复的问题，而直接用set存储就方便许多了),复杂度为 (O(n^2)),如果用set的话，复杂度应该是 (O(n^3))

class Solution {
public:
    vector<vector<int>> threeSum(vector<int>& nums) {
        int i,lo,hi;
        int len = nums.size();
        sort(nums.begin(), nums.end());
        vector<vector<int>> res;
        
        for(int i = 0; i < len-2; i++){
            int lo = i+1,hi = len-1;
            int sum = -nums[i];
            if(i == 0 || (nums[i] != nums[i-1] && i > 0)){
            while(lo < hi){
                if((nums[lo] + nums[hi]) == sum){
                    vector<int> tmp;
                    tmp.push_back(nums[i]);
                    tmp.push_back(nums[lo]);
                    tmp.push_back(nums[hi]);
                    cout << tmp[0] << " " << tmp[1] <<" " << tmp[2] << endl;
                    res.push_back(tmp);
                    while(lo < hi && nums[lo] == nums[lo+1]){
                        lo++;
                    }
                    while(lo < hi && nums[hi] == nums[hi-1]){
                        hi--;
                    }
                    lo++;hi--;
                }
                else if ((nums[lo]+nums[hi]) < sum)
                {
                    lo++;
                }else hi--;
            }
        }
        }
        return res;
    }
};