Problem:

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).

Solution:

此题与15题基本类似，甚至更简单一些，只需要比较和的结果即可，碰到和等于target的时候就直接返回吧！！！

题目大意：

给一个整数数组，找到三个数的和与给定target的值距离最短的那个和

解题思路：

直接看代码把，没看懂的看我的15题题解

Java源代码（用时342ms）：

public class Solution {
    public int threeSumClosest(int[] nums, int target) {
        int length=nums.length,Min=Integer.MAX_VALUE;
        Arrays.sort(nums);
        for(int i=0;i<length-2;i++){
            if(i>0 && nums[i]==nums[i-1])continue;
            int begin=i+1,end=length-1;
            while(begin<end){
                int sum=nums[i]+nums[begin]+nums[end];
                if(Math.abs(sum-target)<Math.abs(Min))Min=sum-target;
                if(sum==target)return target;
                else if(sum>target)end--;
                else begin++;
            }
        }
        return Min+target;
    }
}

C语言源代码（用时9ms）：

int abs(int tar){
    return tar>0?tar:-tar;
}
void quickSort(int* nums,int first,int end){
    int l=first,r=end;
    if(first>=end)return;
    int temp=nums[l];
    while(l<r){
        while(l<r && nums[r]>=temp)r--;
        if(l<r)nums[l]=nums[r];
        while(l<r && nums[l]<=temp)l++;
        if(l<r)nums[r]=nums[l];
    }
    nums[l]=temp;
    quickSort(nums,first,l-1);
    quickSort(nums,l+1,end);
}
int threeSumClosest(int* nums, int numsSize, int target) {
    int begin,end,i,sum,Min=INT_MAX;
    quickSort(nums,0,numsSize-1);
    for(i=0;i<numsSize-2;i++){
        if(i>0 && nums[i]==nums[i-1])continue;
        begin=i+1;end=numsSize-1;
        while(begin<end){
            sum=nums[i]+nums[begin]+nums[end];
            if(abs(sum-target)<abs(Min))Min=sum-target;
            if(sum==target)return target;
            else if(sum>target)end--;
            else begin++;
        }
    }
    return Min+target;
}

C++源代码（用时12ms）:

class Solution {
public:
    int threeSumClosest(vector<int>& nums, int target) {
        int length=nums.size(),Min=2147483647;
        quickSort(nums,0,length-1);
        for(int i=0;i<length-2;i++){
            if(i>0 && nums[i]==nums[i-1])continue;
            int begin=i+1,end=length-1;
            while(begin<end){
                int sum=nums[i]+nums[begin]+nums[end];
                if(abs(sum-target)<abs(Min))Min=sum-target;
                if(sum==target)return target;
                else if(sum>target)end--;
                else begin++;
            }
        }
        return Min+target;
    }
private:
    int abs(int t){
        return t>0?t:-t;
    }
    void quickSort(vector<int>& nums,int first,int end){
        int l=first,r=end,tmp;
        if(first>=end)return;
        tmp=nums[l];
        while(l<r){
            while(l<r && nums[r]>=tmp)r--;
            if(l<r)nums[l]=nums[r];
            while(l<r && nums[l]<=tmp)l++;
            if(l<r)nums[r]=nums[l];
        }
        nums[l]=tmp;
        quickSort(nums,first,l-1);
        quickSort(nums,l+1,end);
    }
};

Python源代码（用时127ms）：

class Solution:
    # @param {integer[]} nums
    # @param {integer} target
    # @return {integer}
    def threeSumClosest(self, nums, target):
        length=len(nums);Min=2147483647
        nums.sort()
        for i in range(length-2):
            if i>0 and nums[i]==nums[i-1]:continue
            begin=i+1;end=length-1
            while begin<end:
                sum=nums[i]+nums[begin]+nums[end]
                if abs(sum-target)<abs(Min):Min=sum-target
                if sum==target:return target
                elif sum>target:end-=1
                else:begin+=1
        return Min+target