Leetcode-Sum, 3Sum, 3Sum Closest, 4Sum

Given an array of integers, return indices of the both numbers such that they add-to a specific target.

You may assume this each input would has exactly one solution, and you could not use the same element twice.

Example:

Given nums = [2, 7, one, +], target = 9,

Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].

Class Solution {public
:
    vector<int> twosum (vector<int>& nums, int target) {
        map<int, int>mp;
        vector<int>res;
        int len = Nums.size ();
        for (int i=0;i<len;i++)
            mp[nums[i]]=i;
        for (int i=0;i<len;i++)
        {
            if (Mp[target-nums[i]])
            {
                res.push_back (i);
                Res.push_back (Mp[target-nums[i]);
                return res;}}}
;

Given an array S of n integers, is 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]] 

Class Solution {public:vector<vector<int>> threesum (vector<int>& nums) {int len=nums.
        Size ();
        vector<vector<int> > Res;
        if (len<=2) return res;
        Sort (Nums.begin (), Nums.end ());
        for (int i=0;i<len;)
            {int st=i+1,ed=len-1; while (st<ed) {if (nums[i]+nums[st]+nums[ed]==0) {RES.P
                    Ush_back ({nums[i],nums[st],nums[ed]});
                    st++;
                    ed--;
                    while (st<ed&&nums[st]==nums[st-1]) st++;
                while (st<ed&&nums[ed]==nums[ed+1]) ed--;
                    } else if (nums[i]+nums[st]+nums[ed]<0) {st++;
                while (st<ed&&nums[st]==nums[st-1]) st++;
} else if (nums[i]+nums[st]+nums[ed]>0) {ed--;                    while (st<ed&&nums[ed]==nums[ed+1]) ed--;
            }} i++;
        while ((I<len) && (nums[i]==nums[i-1)) i++;
    } return res; }
};

Given an array S of n integers, find three integers in S such, the sum was closest to a Given number, target. Return the sum of the three integers. You may assume this each input would has exactly one solution.

    For example, given array S = {-1 2 1-4}, and target = 1.

    The sum is closest to the target is 2. (-1 + 2 + 1 = 2).

class solution {public:int threesumclosest (vector<int>& nums, int target) {int L
        En=nums.size ();
        Sort (Nums.begin (), Nums.end ());
        int res=nums[0]+nums[1]+nums[2];
        if (len==3) return res;
            for (int i=0;i<len-2;i++) {int st=i+1,ed=len-1;
                while (st<ed) {int sum=nums[i]+nums[st]+nums[ed];
                    if (ABS (Sum-target) <abs (res-target)) {res=sum;
                if (res==target) return res;
                } if (Sum>target) ed--;
            else st++;
    }} return res; }
};

Given an array S of n integers, is there elements a, B, C, and D in S such that A + B + c + d = target? Find all unique quadruplets in the array which gives the sum of target.

Note:the solution set must not contain duplicate quadruplets.

For example, given array S = [1, 0,-1, 0,-2, 2], and target = 0.

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

Class Solution {public:vector<vector<int>> foursum (vector<int>& nums, int target) {ve
        ctor<vector<int>> Res;
        int len=nums.size ();
        if (len<4) return res;
        Sort (Nums.begin (), Nums.end ());
            for (int i=0;i<len-3;i++) {if (i>0&&nums[i]==nums[i-1]) continue;
            if (nums[i]+nums[i+1]+nums[i+2]+nums[i+3]>target) break;
            if (nums[i]+nums[len-1]+nums[len-2]+nums[len-3]<target) continue;
                for (int j=i+1;j<len-2;j++) {if (j>i+1&&nums[j]==nums[j-1]) continue;
                if (nums[i]+nums[j]+nums[j+1]+nums[j+2]>target) break;
                if (nums[i]+nums[j]+nums[len-1]+nums[len-2]<target) continue;
                int st=j+1,ed=len-1;
                    while (st<ed) {int sum=nums[i]+nums[j]+nums[st]+nums[ed];
if (sum==target) {                        Res.push_back ({nums[i],nums[j],nums[st],nums[ed]});
                        ed--;
                        while (st<ed&&nums[ed]==nums[ed+1]) ed--;
                        st++;
                    while (st<ed&&nums[st]==nums[st-1]) st++;
                        } else if (sum>target) {ed--;
                    while (st<ed&&nums[ed]==nums[ed+1]) ed--;
                        } else if (sum<target) {st++;
                    while (st<ed&&nums[st]==nums[st-1]) st++;
    }}}} return res; }
};