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).
thinking:
(1)是不是很熟悉,这道题只不过是求三数和为一个特定值的变形
(2)仍然采用 贪心的思想,三指针二分查找,寻找三数和与target差值的绝对值最小的和数,这里采用了flag标记正负关系。
code :
class Solution { public: int threeSumClosest(vector<int> &num, int target) { int index; bool flag=true; sort(num.begin(),num.end()); if(num.at(0)+num.at(1)+num.at(2)>target) index=num.at(0)+num.at(1)+num.at(2)-target ; else { index=target-(num.at(0)+num.at(1)+num.at(2)); flag=false; } for (int i = 0; i < num.size(); ++i) { int p = i + 1, q = num.size() - 1; int sum=0; while (p < q) { sum = num[i] + num[p] + num[q]; if (sum == target) { return sum; }//if else if (sum < target) //和太小,p向后移动 { ++p; if(target-sum<index) { index=target-sum; flag=false; } } else //和过大,q向前移动 { --q; if(sum-target<index) { index=sum-target; flag=true; } }//else }//while }//for if(flag) return index+target; else return target-index; } };<strong> </strong>
leetcode 题解 || 3Sum Closest 问题
原文地址:http://blog.csdn.net/hustyangju/article/details/44452895