LeetCode 16 3Sum Closest(C,C++,Java,Python)，leetcode3sum

LeetCode 16 3Sum Closest(C,C++,Java,Python)，leetcode3sum
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:此題與15題基本類似，甚至更簡單一些，只需要比較和的結果即可，碰到和等於target的時候就直接返回吧！！！
題目大意：給一個整數數組，找到三個數的和與給定target的值距離最短的那個和
解題思路：直接看代碼把，沒看懂的看我的15題題解
Java原始碼（用時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語言原始碼（用時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++原始碼（用時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原始碼（用時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