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).
算法:固定第一个,另外两个数适用双指针。O(N^2)
java:
public class Solution { public int threeSumClosest(int[] num, int target) { int len=num.length; int distance=Integer.MAX_VALUE; int sum=0; Arrays.sort(num); for(int i=0;i<len-2;i++){ int s=i+1; int e=len-1; while(s<e){ int cnt= num[i]+num[s]+num[e]; int dis = Math.abs(target-cnt); if(dis<=distance){ sum=cnt; distance=dis; if(sum==target){ return sum; }else if(sum<target){ s++; }else { e--; } }else if(cnt<target){ s++; }else{ e--; } } } return sum; } }
c++:
class Solution { public: int threeSumClosest(vector<int> &num, int target) { int size=num.size(); sort(num.begin(),num.end()); int close=0; int dis=INT_MAX; for(int i=0;i<size-2;i++){ int j=i+1; int k=size-1; while(j<k){ int sum=num[i]+num[j]+num[k]; if(abs(sum-target)<=dis){ dis=abs(sum-target); close=sum; if(sum==target){ return close; }else if(sum<target){ j++; }else{ k--; } }else if(sum-target<0){ j++; }else{ k--; } } } return close; } };
原文地址:http://blog.csdn.net/u010786672/article/details/41148591