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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).
思路：
先排序，然后固定一个边界，另外两个边界收缩。
class Solution { public: int threeSumClosest(vector<int> &num, int target) { int l, r, m; int ans = num[0] + num[1] + num[2]; sort(num.begin(), num.end()); //数字从小到大排序 for(l = 0; l < num.size() - 2; l++) //固定左边界 { m = l + 1; r = num.size() - 1; while(m < r) //收缩中间和右边界 { int tmp = num[l] + num[m] + num[r]; ans = (abs(ans - target) > abs(tmp - target)) ? tmp : ans; if(tmp < target) { (num[m] > num[r]) ? r-- : m++; //比目标小 则把小的数字变大 } else if(tmp > target) { (num[m] < num[r]) ? r-- : m++; //比目标大 则把大的数字变小 } else { return tmp; //已经等于target就直接返回 } } } return ans; } };
【leetcode】3Sum Closest（middle）
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原文地址：http://www.cnblogs.com/dplearning/p/4276135.html
