leetcode(M) 3sum closest

1. 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).

代码：

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

后来看到leetcode讨论区，这个应该再加上判断nums.size()<=3,return accumulate(nums.begin(),nums.end(),0), 其中accumulate是用于累计计算的函数，默认情况下是求和。而其中第三个参数便是和的初始值。如果数据为a0, …, an，则accumulate返回a0 + … + an + 初始值。
同时if(A-target<0) ++j; if(A-target>0) --k;
可以简化成 A-target<0?++j:--k;

代码最终版：

class solution{public:int threeSumClosest(vector<int>& nums, int target){if(nums.size()<=3) return accumulate(nums.begin(),nums.end(),0);sort(nums.begin(),nums.end());int i,j,k,A,res=nums[0]+nums[1]+nums[2];for(i=0;i<nums.size();++i){j=i+1;k=nums.size()-1;while(j<k){A=nums[i]+nums[j]+nums[k];res=abs(res-target)<abs(A-target)?res:A;(A-target)>0?--k:++j;}}return res;}};