Leetcode 16. 3sum closest

@author stormma
@date 2017/11/03

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生命不息，奋斗不止

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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;    }}