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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).
求三个数的和sum，使sum与target相差最小。这题实际上跟前面坐的2Sum，以及3Sum相差不是很大，大体的想法是先排序，然后外层的for循环遍历从小到大的数，然后从剩下的数中取三数之和，然后把最小的一步步保存下来，最终将最小的返回就行了。代码如下：
1 class Solution { 2 public: 3 int threeSumClosest(vector<int>& nums, int target) { 4 int sz = nums.size(); 5 sort(nums.begin(), nums.end());//当然了，首先还是要排序的 6 int diff = INT_MAX;//这个直接取INT_MAX,这样计算出来的diff一定是比这个要小的 7 int tmpDiff; 8 int ans; 9 int beg, end; 10 for (int i = 0; i < sz - 2; ++i){ 11 beg = i + 1, end = sz - 1; 12 while (beg < end){ 13 int sum = nums[i] + nums[beg] + nums[end]; 14 if ((tmpDiff = abs(sum - target)) < diff){ 15 diff = tmpDiff; 16 ans = sum; 17 } 18 if (sum > target){ 19 end--; 20 } 21 else if (sum < target){ 22 beg++; 23 } 24 else 25 return sum;//相等的话就直接返回了 26 } 27 } 28 return ans; 29 } 30 };
大体上就是这样
LeetCode OJ：3Sum Closest（最接近的三数之和）
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原文地址：http://www.cnblogs.com/-wang-cheng/p/4857149.html
