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).
1 class Solution { 2 public: 3 int threeSumClosest(vector<int>& num, int target) { 4 int result = 0; 5 int min_gap = INT_MAX; 6 sort(num.begin(), num.end()); 7 for (size_t i = 0; i < num.size() - 2; ++i) { 8 if (i > 0 && num[i] == num[i - 1]) continue; 9 size_t j = i + 1, k = num.size() - 1; 10 while (j < k) { 11 int sum = num[i] + num[j] + num[k]; 12 int gap = abs(sum - target); 13 if (gap < min_gap) { 14 min_gap = gap; 15 result = sum; 16 } 17 if (sum < target) { 18 ++j; 19 } else if (sum > target) { 20 --k; 21 } else { 22 return result; 23 } 24 } 25 } 26 return result; 27 } 28 };
与上题3Sum类似，先排序，然后以一个元素为准，两边夹逼，逐个求三个元素的和，找到与目标相等的了可以直接返回，对于重复元素可以只找一次。
【Leetcode】3Sum Closest,布布扣,bubuko.com
原文：http://www.cnblogs.com/dengeven/p/3606227.html
