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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).
分析：同样采取i,j,k三个对数组扫描，记录target closet to target;
code:
class Solution {public: int threeSumClosest(vector<int> &num, int target) { int subnum=num[0]+num[1]+num[2]; int sum; sort(num.begin(),num.end()); for(int i=0;i<num.size();++i) { int j=i+1; int k=num.size()-1; while(j<k) { sum=num[i]+num[j]+num[k]; if(sum==target) return target; else if(sum>target) { if(abs(sum-target)<abs(subnum-target)) subnum=sum; --k; } else { if(abs(sum-target)<abs(subnum-target)) subnum=sum; ++j; } } int tmp=abs(sum-target); if(tmp<abs(subnum-target)) subnum=sum; } return subnum; }};
上面的程序比较繁琐，可以简化
code：
class Solution {public: int threeSumClosest(vector<int> &num, int target) { int subnum=num[0]+num[1]+num[2]; int sum; sort(num.begin(),num.end()); for(int i=0;i<num.size();++i) { int j=i+1; int k=num.size()-1; while(j<k) { sum=num[i]+num[j]+num[k]; if(abs(sum-target)<abs(subnum-target)) subnum=sum; if(sum>target) --k; else ++j; } } return subnum; }};
3Sum Closest
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