第一題、15. 3Sum
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]
]
二分查詢的思想

vector<vector<int> > threeSum(vector<int> &num) {
        vector<vector<int> > ret;
        int size = num.size();
        sort(num.begin(), num.end());
        for(int i = 0; i < size; i ++)
        {
            //skip same i
            while(i > 0 && i < size && num[i] == num[i-1
 
])
                i ++;
            int j = i + 1;
            int k = size - 1;
            while(j < k)
            {
                int sum = num[i] + num[j] + num[k];
                if(sum == 0)
                {
                    vector<int> cur(3);
                    cur[0] = num[i];
                    cur[1
 
] = num[j];
                    cur[2] = num[k];
                    ret.push_back(cur);
                    j ++;
                    k --;
                    //skip same j
                    while(j < k && num[j] == num[j-1])
                        j ++;
                    //skip same k
                    while(k > j && num[k] == num[k+1])
                        k --;
                }
                else if(sum < 0)
                {
                    j ++;

                }
                else
                {
                    k --;

                }
            }
        }
        return ret;
    }

第二題：16. 3Sum Closest
Given an array S of n integers, find three integers in S such that the sum is closest to a given number, target. Return the sum of the three integers. You may assume that each input would have exactly one solution.
For example, given array S = {-1 2 1 -4}, and target = 1.
The sum that is closest to the target is 2. (-1 + 2 + 1 = 2).

int threeSumClosest(vector<int>& nums, int target) {
        vector<vector<int> > ret;
        int size = nums.size();
        sort(nums.begin(), nums.end());
        int sum = nums[0]+nums[1]+nums[size-1];
        int minDis = sum - target;
        for(int i = 0; i < size; i++)
        {
            int j = i + 1;
            int k = size -1;
            while(j<k)
            {
                sum = nums[i]+nums[j]+nums[k];
                if(abs(sum - target) < abs(minDis))
                {
                    minDis = sum - target;
                }

                //minDis = sum - target;
                if(sum == target)//增加這部分處理，是對於{1,2,3,4,5,6,7} target=10,這種一旦找到1,2,7三個數，則不用繼續遍歷，繼續
                {
                    return target;
                }
                else if(sum > target)
                {
                    k--;
                }
                else
                {
                    j++;
                }
            }
        }

        return minDis + target;
    }

第三題、18. 4Sum
Given an array S of n integers, are 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]
]

/*
    對陣列排序
    確定四元數中的前兩個（a，b）
    遍歷剩餘陣列確定兩外兩個（c，d），確定cd時思路跟3Sum確定後兩個資料一樣，二分查詢左右逼近。
    在去重時採用set集合
    */
    vector<vector<int>> fourSum(vector<int>& nums, int target) {
        vector<vector<int>> result;
        set<vector<int>> setRes;
        int len = nums.size();
        sort(nums.begin(),nums.end());
        if(len < 4)
        {
            return (vector<vector<int>> ());
        }

        /*
        for(int i = 0; i < len; i++)
        {
            for(int j = i + 1; j < len; j++)
        */
        for(int i = 0; i < len - 3; i++)
        {
            for(int j = i + 1; j < len - 2; j++)
            {
                //二分查詢
                int begin = j + 1;
                int end = len -1;
                while(begin < end)
                {

                    int sum = nums[i] + nums[j] + nums[begin] + nums[end];
                    if(sum == target)
                    {
                        vector<int> temp;
                        temp.push_back(nums[i]);
                        temp.push_back(nums[j]);
                        temp.push_back(nums[begin]);
                        temp.push_back(nums[end]);
                        setRes.insert(temp);
                        begin++;
                        end--;
                    }
                    else if(sum > target)
                    {
                        end--;
                    }
                    else
                    {
                        begin++;
                    }
                }
            }
        }

        set<vector<int>>::iterator iter;
        for(iter = setRes.begin(); iter!=setRes.end(); iter++)
        {
            result.push_back(*iter);
        }

        return result;

    }

時間複雜度更低的程式碼：

vector<vector<int>> fourSum(vector<int>& nums, int target) {
        vector<vector<int>> total;
        int n = nums.size();
        if(n<4)  return total;
        sort(nums.begin(),nums.end());
        for(int i=0;i<n-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[n-3]+nums[n-2]+nums[n-1]<target) continue;
            for(int j=i+1;j<n-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[n-2]+nums[n-1]<target) continue;
                int left=j+1,right=n-1;
                while(left<right){
                    int sum=nums[left]+nums[right]+nums[i]+nums[j];
                    if(sum<target) left++;
                    else if(sum>target) right--;
                    else{
                        total.push_back(vector<int>{nums[i],nums[j],nums[left],nums[right]});
                        do{left++;}while(nums[left]==nums[left-1]&&left<right);
                        do{right--;}while(nums[right]==nums[right+1]&&left<right);
                    }
                }
            }
        }
        return total;
    }