LeetCode 16. 3Sum Closest LeetCode 18. 4Sum

题目链接：https://leetcode.com/problems/3sum-closest/

题目描述：

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).

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思路：参考链接：http://blog.csdn.net/doc_sgl/article/details/12462151

    参考2sum问题的解法，降低算法复杂度的关键是先排好序，然后设置首尾设置两个指针，当sum>target时，尾部指针--，反之首部指针++，进行线性搜索。 本题不是sum==target，那么只需不断更新最小值。

一开始写的暴力搜索的三个，都超时了，不过功能上应该没有什么问题，测试简单的情况通过了，应该只是时间复杂度比较高。。。

方法一：

class Solution {public:    int threeSumClosest(vector<int>& nums, int target) {        vector<int> sum;        int result;        int len1=nums.size();        for(int i=0;i<len1;i++)        {            for(int j=i+1;j<len1;j++)            {                for(int k=j+1;k<len1;k++)                {                    sum.push_back(nums[i]+nums[j]+nums[k]);                }            }        }        int len2=sum.size();        int index=0;        int minSum=abs(sum[0]-target);        for(int m=0;m<len2-1;m++)        {            if(abs(sum[m+1]-target)<minSum)            {                minSum=abs(sum[m+1]-target);                index=m+1;            }        }        result=sum[index];        return result;    }};

方法二：

class Solution {public:    int threeSumClosest(vector<int>& nums, int target) {        vector<int> sum;        int result;        int len1=nums.size();        sum.push_back(nums[0]+nums[1]+nums[2]);        int minSum=abs(sum[0]-target);        int index=0;        int flag=0;        int m=0;        for(int i=0;i<len1;i++)        {            for(int j=i+1;j<len1;j++)            {                for(int k=j+1;k<len1;k++)                {                    flag=0;                    if(i+j+k==1)                    {                        flag=1;                           break;                    }                    sum.push_back(nums[i]+nums[j]+nums[k]);                    m++;                    if(abs(sum[m]-target)<minSum)                    {                        minSum=abs(sum[m]-target);                        index=m;                    }                }                if(flag==1)                    break;            }            if(flag==1)                break;        }        result=sum[index];        return result;    }};

方法三：

class Solution {public:int threeSumClosest(vector<int>& nums, int target) {vector<int> sum;int result;int len1 = nums.size();int m = 0;int tmp;map<int, int> dif2sumindex;map <int, int>::iterator beg;int index;for (int i = 0; i<len1; i++){for (int j = i + 1; j<len1; j++){for (int k = j + 1; k<len1; k++){tmp = nums[i] + nums[j] + nums[k];sum.push_back(tmp);dif2sumindex[abs(tmp - target)] = m;m++;}}}beg = dif2sumindex.begin();index=beg->second;result = sum[index];return result;}};

方法四：(AC的算法)

class Solution {public:int threeSumClosest(vector<int>& nums, int target) {        sort(nums.begin(),nums.end());        int len=nums.size();        int i=1,j=len-1;        int minDist=abs(nums[0]+nums[1]+nums[len-1]-target);        int result=nums[0]+nums[1]+nums[len-1];        int tmp=0;        for(int k=0;k<len-2;k++)        {            if(k>0&&nums[k]==nums[k-1])                continue;            i=k+1;j=len-1;            while(i<j)            {                tmp=nums[k]+nums[i]+nums[j];                if(abs(tmp-target)<minDist)                {                    minDist=abs(tmp-target);                    result=tmp;                }                if(tmp==target)                    return target;                else if(tmp-target>0)                    j--;                else                    i++;            }        }        return result;}};

题目链接：https://leetcode.com/problems/4sum/

题目描述：

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]]

思路：参考前面3sum问题的思路，先用两重for循环确定前两个数字，在剩下的数里面用双指针用线性搜索，算法时间复杂度：O(n3)，还有待于改进（可能会用到哈希表，还不太熟悉，以后来改进）

class Solution {public:vector<vector<int>> fourSum(vector<int>& nums, int target) {vector<vector<int>> re;sort(nums.begin(),nums.end());map<vector<int>,int> mark;int n = nums.size();for (int i = 0; i<n - 3; i++){for (int j = i + 1; j< n - 2; j++){int p1 = j + 1, p2 = n - 1;int sums = nums[i] + nums[j];while (p1<p2){if (sums + nums[p1] + nums[p2] == target){vector<int> tmp = { nums[i], nums[j], nums[p1], nums[p2] };    if(mark[tmp]==0)    {    re.push_back(tmp);    mark[tmp]=1;    }p1++;p2--;}else if (sums + nums[p1] + nums[p2]<target)p1++;elsep2--;}}}return re;}};