[JavaScript 刷题] 数组 - 划分数组使最大差为 K,leetcode 2294

Posted 2022-06-11 GoldenaArcher

[javascript 刷题] 数组 - 划分数组使最大差为 K,leetcode 2294

github repo 地址: https://github.com/GoldenaArcher/js_leetcode，Github 的目录 大概 会更新的更勤快一些。

题目地址：2294. Partition Array Such That Maximum Difference Is K

题目

如下：

You are given an integer array nums and an integer k. You may partition nums into one or more subsequences such that each element in nums appears in exactly one of the subsequences.

Return the minimum number of subsequences needed such that the difference between the maximum and minimum values in each subsequence is at most k.

A subsequence is a sequence that can be derived from another sequence by deleting some or no elements without changing the order of the remaining elements.

解题思路

基本上说看到这种题，包括最大最小值，第一个考虑的就是排序。

以这道题的例子来说：

nums = [3,6,1,2,5], k = 2

We can partition nums into the two subsequences [3,1,2] and [6,5].

对于这道题的 subsequence 来说，顺序就不是这么的重要了，如果将这个数组进行排序之后就能很方便的解决这道题。

[3,6,1,2,5] 进行排序后就能获得 [1,2,3,5,6]，这时候只要保存一个变量作为当前 subsequence 中的最小值，随后遍历到 nums[i + n] - nums[i] > k 的点再重复该步骤即可。

使用 JavaScript 解题

/**
 * @param number[] nums
 * @param number k
 * @return number
 */
var partitionArray = function (nums, k) 
  nums.sort((a, b) => a - b);

  let counter = 0;
  for (let i = 0; i < nums.length; i++) 
    const curr = nums[i];
    while (nums[i] - curr <= k) 
      i++;
    
    i--;
    counter++;
  

  return counter;
;

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