"Intermediate algorithm" 1. Three number of the sum

Topic:

0= [-1012,-1,-4], the ternary set that satisfies the requirement is: [  [ -101],  [-1,-12]] 

Problem Solving Ideas:

classSolution { Public:/*vector<vector<int>> threesum (vector<int>& nums) {sort (Nums.begin (), Nums.end ());        vector<vector<int>> Res;                int len = Nums.size ();        if (Nums.size () < 3) {return res; } for (int i = nums.size () -1;i >=2;--i) {for (int j = i-1;j >= 1;--j) {for (in  t k = j-1;k >=0;--k) {if (Nums[i] + nums[j] + nums[k] = = 0) {vector<int>                        tmp                        Tmp.push_back (Nums[i]);                        Tmp.push_back (Nums[j]);                        Tmp.push_back (Nums[k]);                    Res.push_back (TMP);    }}}} return res; }*/    intBinarySearch (vector<int>& Nums,intTargetintStartintend) {        if(Start >end) {            return-1; }                 while(Start <=end) {            intMid = (start + end)/2; if(Nums[mid] = =target) {                returnmid; }            if(Nums[mid] >target) {End= Mid-1; }            if(Nums[mid] <target) {Start= Mid +1; }        }                return-1; }        BOOLFindtree (vector<vector<int>> Nums,intAintBintc) {         for(inti =0; i < nums.size (); ++i) {            if(nums[i][0] = = a &&nums[i][1] = = B &&nums[i][2] ==c) {                return true; }        }                return false; } Vector<vector<int>> Threesum (vector<int>&nums) {Vector<vector<int>>Res; intLen =nums.size (); inttarget =0; if(Len <3){            returnRes;                } sort (Nums.begin (), Nums.end ());  for(inti =0; I < len-2; ++i) {            if(I >0&& nums[i-1] ==Nums[i]) {                Continue; }             for(intj = i +1, k = Len-1; J <K;) {if(J > i +1&& Nums[j] = = nums[j-1]) {J++; Continue; }                if(K < Len-1&& Nums[k] = = nums[k+1]) {k--; Continue; }                intsum = Nums[i] + nums[j] +Nums[k]; if(Sum >target) {k--; }Else if(Sum <target) {J++; }Else{vector<int>tmp;                    Tmp.push_back (Nums[i]);                    Tmp.push_back (Nums[j]);                    Tmp.push_back (Nums[k]);                    Res.push_back (TMP); J++; K--; }            }        }                returnRes; }};

"Intermediate algorithm" 1. Three number of the sum