leetcode 16 3Sum Closet

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3Sum Closest
Total Accepted: 53147 Total Submissions: 195388 Difficulty: Medium

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).

这道题的思路和之前的4Sum的思路一致，先把所有数增序排序，然后用三个指针，第一个指针从头到尾走形成第一层循环，第二个指针从第一个指针的下一个开始往后走，第三个指针从最后一个开始往前走，当第二个指针和第三个指针相遇时，就穷尽了在第一个指针固定的情况下，第二个数和第三个数的可能产生最接近值的组合情况，采用的策略是如果第二个数和第三个数相加的和大于target，则第三个指针往前走一个，因为之前已经按增序排序，所以第三个指针指向一个比之前小的值，如果第二个数和第三个数相加的和小于target，则第二个指针往后走一个，所以第二个指针指向一个比之前大的值。所以这个算法实际上是一个双重循环。

下面是完整程序代码：

#include <iostream>#include <stdlib.h>using namespace std;int abs(int n){    return n>0?n:-1*n;}int cmp(const void *a, const void *b){    return *(int*)a - *(int*)b;}int threeSumClosest(int* nums, int numsSize, int target) {    qsort(nums, numsSize, sizeof(int), cmp);    /*for(int i = 0; i < numsSize; i++)    {        cout<<nums[i]<<" ";    }*/    int optimal, difference = 10000000, p, q, tmp;    for(int i = 0; i < numsSize - 2; i++)    {        p = i + 1;        q = numsSize - 1;        while(p < q)        {            tmp = nums[i] + nums[p] + nums[q] - target;            if(abs(tmp) < difference)            {                difference = abs(tmp);                optimal = nums[i] + nums[p] + nums[q];            }            if(tmp < 0)                p++;            else if(tmp == 0)                return optimal;            else                q--;        }    }    return optimal;}int main(){    int numsSize, *nums, target;    cout<<"Input the size of numbers:"<<endl;    cin>>numsSize;    nums = (int*)malloc(sizeof(int) * numsSize);    for(int i = 0; i < numsSize; i++)    {        cin>>nums[i];    }    cout<<"Input the target:"<<endl;    cin>>target;    cout<<endl<<threeSumClosest(nums, numsSize, target)<<endl; }

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