Question

我正在看problem solution on leetcode，并且有一个聪明的解决方案，它使用Python的map函数将字符串列表转换为相同大小的矩阵，如下所示：

t = map(None, *words)

解释：

地图（无，...）转置“矩阵”，用无填充缺失的点。例如：

["abc",           [('a', 'd', 'f'),
 "de",     =>      ('b', 'e', None),
 "f"]              ('c', None, None)]

我想知道我是否可以用 js 要点达成类似的目标

Answer 1

创建矩阵（长度相等）：

solution=array.reduce((solution,el)=>(el.split("").forEach((letter,index)=>(solution[index]=solution[index]||[]).push(letter)),solution),[]);
//if you really need the *undefineds* :
maxlength=array.reduce((length,arr)=>Math.max(length,arr.length),0);
solution.forEach(el=>el.length=maxlength);

http://jsbin.com/nisoderini/edit?console 由于某种原因而留下的解释...

检查数组中的等长数组：

length=array.reduce((length,arr,i)=>!i?arr.length:(length?(arr.length==length?length:false):false),0);

如果数组不是对称矩阵，则长度为假...

检查方阵

square=array.every(arr=>arr.length==array.length);

Answer 2

我就是这样做的。

function transpose(mat, replacement){
    const len = findMaxArr(mat);
    const res = [];
    // loop through each row...
    mat.map(e => {
        const currElemArr = e.split('')
        if (e.length === len) {
            res.push(currElemArr);
        } else {
            // if current row len is less than max len
            // pad it with the replacement variable
            const diff = len - e.length;
            const currElemArrPadded = currElemArr.concat(createRepArr(replacement, diff));
            res.push(currElemArrPadded);
        }
    });
    return res;
}

// create an array of inp of length len
function createRepArr(inp, len) {
    const res = [];
    for (let i = 0; i < len; i++) {
        res.push(inp);
    }
    return res;
}

// find maximum length array of string in a string array
function findMaxArr(mat) {
    let res = 0;
    mat.forEach(m => {
        if (m.length > res) {
            res = m.length;
        }
    });
    return res;
}

const data = ["abc", "de", "f"];
console.log(transpose(data, 0));

JS - 将字符串列表映射到等长矩阵的有效方法

2 个答案: