leetcode 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).

题意就是说寻找三个数据之和距离target最近的，思路和上一道题3Sum完全一样，只是修改一下寻找的目标即可，也即对于3Sum，先确定一个数字，然后这个问题就退化成了2Sum的问题。针对2Sum，先对数组排序，然后使用双指针匹配可行解就可以解决。
代码如下：

import java.util.Arrays;public class Solution {    int minLen=Integer.MAX_VALUE;    int finsum=0;    public int threeSumClosest(int[] nums, int target)    {        if(nums==null || nums.length<3)            return 0;        if(nums.length==3)            return nums[0]+nums[1]+nums[2];        Arrays.sort(nums);        for(int i=0;i<nums.length;i++)        {            //重复的数据就不在统计            if(i>=1 && nums[i]==nums[i-1])                continue;            //System.out.println("Index: "+nums[i]);            getRes(nums,i+1,nums[i],target);        }        return finsum;    }    private void getRes(int[] nums, int beg,int a,int target)    {        int i=beg,j=nums.length-1;        while(i<j)        {            int sum=nums[i]+nums[j]+a;            if(sum > target)            {                if(minLen > sum-target)                {                    minLen=sum-target;                    finsum=sum;                }                j--;            }else if(sum < target)            {                if(minLen > target-sum)                {                    minLen=target-sum;                    finsum=sum;                 }                i++;            }else             {                minLen=target-sum;                finsum=sum;                 break;            }        }    }    public static void main(String[] args)    {        Solution so=new Solution();        int []a={1,1,1,1};        System.out.println(so.threeSumClosest(a,3));    }}

C++做法如下，主要就是直接设置一个比较即可。

代码如下：

#include <iostream>#include <vector>#include <algorithm>#include <climits>using namespace std;class Solution{public:    int finalSum = 0;    int minLen = numeric_limits<int>::max();    int threeSumClosest(vector<int>& nums, int target)    {        if (nums.size() < 3)            return 0;        else if (nums.size() == 3)            return nums[0] + nums[1] + nums[2];        sort(nums.begin(), nums.end());        for (int i = 0; i < nums.size(); i++)        {            if (i >= 1 && nums[i] == nums[i - 1])                continue;            dealWithTwoSum(nums, i + 1, nums[i],target);        }        return finalSum;    }    void  dealWithTwoSum(vector<int>& nums, int beg, int now,int target)    {        int i = beg, j = nums.size() - 1;        while (i < j)        {            int sum = nums[i] + nums[j] + now;            if (sum > target)            {                if (minLen > sum - target)                {                    minLen = sum - target;                    finalSum = sum;                }                j--;            }            else if (sum < target)            {                if (minLen > target - sum)                {                    minLen = target - sum;                    finalSum = sum;                }                i++;            }            else            {                minLen = 0;                finalSum = sum;                break;            }        }    }};