15. 3Sum，16. 3Sum Closest，18. 4Sum（最后一个方法重要）重要

第一题、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]
]
二分查找的思想

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;    }

第二题：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)//增加这部分处理，是对于{1,2,3,4,5,6,7} target=10,这种一旦找到1,2,7三个数，则不用继续遍历，继续                {                    return target;                }                else if(sum > target)                {                    k--;                }                else                {                    j++;                }            }        }        return minDis + target;    }

第三题、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]
]

/*    对数组排序    确定四元数中的前两个（a，b）    遍历剩余数组确定两外两个（c，d），确定cd时思路跟3Sum确定后两个数据一样，二分查找左右逼近。    在去重时采用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++)            {                //二分查找                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;    }

时间复杂度更低的代码：

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;    }