Leetcode 16. 3sum closest
2017-11-03 10:55
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@author stormma
@date 2017/11/03
生命不息,奋斗不止
先确定1一个数,two pointer去查找,查找之后和target比较,比较上一次(sum - target) 和这次的(sum - target)的差值(sum表示我们查找的三个数的和)
代码实现
@date 2017/11/03
生命不息,奋斗不止
题目
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).
思路分析
其实就是3sum的翻版,只要掌握了上一个题目的做法,这个理所当然可以套上一套题目的做法。先确定1一个数,two pointer去查找,查找之后和target比较,比较上一次(sum - target) 和这次的(sum - target)的差值(sum表示我们查找的三个数的和)
代码实现
class Solution { public int threeSumClosest(int[] nums, int target) { if (nums.length == 1) { return nums[0]; } Arrays.sort(nums); int result = Integer.MAX_VALUE >> 1; for (int i = 0; i < nums.length - 2; i++) { int low = i + 1, high = nums.length - 1; int goal = target - nums[i]; while (low < high) { if (goal == nums[low] + nums[high]) { return target; } result = Math.abs(goal - nums[low] - nums[high]) > Math.abs(result - target) ? result: nums[i] + nums[low] + nums[high]; if (goal > nums[low] + nums[high]) { low++; } else { high--; } } } return result; } }
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