Javascript将数组中的数字与一系列连续数字分组 [英] Javascript group the numbers from an array with series of consecutive numbers

问题描述

例如给定一个已排序的整数数组

a = [0,1,2,5,6,9];

我想确定范围，如

<预><代码>[[0,1,2],[5,6],[9]]

到目前为止，我已经尝试了双重/三重循环，但它嵌套了非常讨厌的代码.也许这个问题可以使用递归或其他聪明的技巧来解决?

<小时>附加示例:

输入

b = [0,1,5,6,7,9];

输出

<预><代码>[[0,1],[5,6,7],[9]]

解决方案

用 Array#reduce 迭代，每当最后一个数字不等于新数字 - 1 时，添加另一个子数组.将当前数字添加到最后一个子数组:

const a = [0,1,2,5,6,9];const 结果 = a.reduce((r, n) => {const lastSubArray = r[r.length - 1];if(!lastSubArray || lastSubArray[lastSubArray.length - 1] !== n - 1) {r.push([]);}r[r.length - 1].push(n);返回 r;}, []);console.log(result);

Given a sorted array of ints for example

a = [0,1,2,5,6,9];

I would like to identify the ranges like

[
    [0,1,2],
    [5,6],
    [9]
]

So far I have tried a double/triple loop but it nests to really nasty code. Maybe this problem can be solved using recursion or other smart tricks?

---

Additional example:

input

b = [0,1,5,6,7,9];

output

[
    [0,1],
    [5,6,7],
    [9]
]

解决方案

Iterate with Array#reduce, and whenever the last number is not equal to the new number - 1, add another sub array. Add the current number to the last sub array:

const a = [0,1,2,5,6,9];

const result = a.reduce((r, n) => {
  const lastSubArray = r[r.length - 1];
  
  if(!lastSubArray || lastSubArray[lastSubArray.length - 1] !== n - 1) {
    r.push([]);
  } 
  
  r[r.length - 1].push(n);
  
  return r;  
}, []);

console.log(result);

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