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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).
相关题目：2sum，3sum
sum < target 能直接 ++left 是因为--right 会 扩大其与target 的距离。
注意：防止溢出。
1 public class Solution { 2 public int threeSumClosest(int[] nums, int target) { 3 Arrays.sort(nums); 4 long closest = Integer.MAX_VALUE; 5 for (int i = 0; i < nums.length - 2; i++) { 6 int left = i + 1; 7 int right = nums.length - 1; 8 int sum = 0; 9 while (left < right) { 10 sum = nums[i] + nums[left] + nums[right]; 11 if (sum == target) { 12 return sum; 13 } else if (sum < target) { 14 ++left; 15 } else { 16 --right; 17 } 18 closest = Math.abs(sum - target) < Math.abs(closest - target) ? sum : closest; 19 } 20 } 21 return (int)closest; 22 } 23 }
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原文地址：http://www.cnblogs.com/FLAGyuri/p/5437472.html
