问题描述:
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).
原问题链接:https://leetcode.com/problems/3sum-closest/
问题分析
这个问题的思路和前面3sum的过程很接近。就是首先对数组排序,然后取target和里面取一个数的差。再到数组里去查找比较和这个差接近的值。如果相等的话就直接返回,否则记录它们的和与target值的偏差,记录下来偏差最小的那个返回。
详细的实现代码如下:
public class Solution { public int threeSumClosest(int[] nums, int target) { Arrays.sort(nums); int result = 0, dif = Integer.MAX_VALUE; for(int i = 0; i < nums.length - 2; i++) { int l = i + 1, r = nums.length - 1; while(l < r) { if(nums[l] + nums[r] == target - nums[i]) return target; else if(nums[l] + nums[r] < target - nums[i]) { if(target - nums[i] - nums[l] - nums[r] < dif) { dif = target - nums[i] - nums[l] - nums[r]; result = nums[l] + nums[r] + nums[i]; } l++; } else { if(nums[l] + nums[r] + nums[i] - target < dif) { dif = nums[l] + nums[r] + nums[i] - target; result = nums[l] + nums[r] + nums[i]; } r--; } } } return result; } }
总体的时间复杂度为O(N * N)。这里要注意的是实现的细节里取的索引值的范围。