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嘗試解決算法問(wèn)題的曲折模式

JavaScript

瀟湘沐 2022-06-09 16:47:02

問(wèn)題如下：字符串“PAYPALISHIRING”在給定數(shù)量的行上以鋸齒形圖案寫入，如下所示：（您可能希望以固定字體顯示此圖案以提高可讀性）P A H NA P L S I I GY I R然后逐行閱讀：“PAHNAPLSIIGYIR”編寫將采用字符串并在給定行數(shù)的情況下進(jìn)行此轉(zhuǎn)換的代碼：字符串轉(zhuǎn)換（字符串 s，int numRows）；示例 1：輸入：s = "PAYPALISHIRING", numRows = 3 輸出："PAHNAPLSIIGYIR" 示例 2：輸入：s = "PAYPALISHIRING", numRows = 4 輸出："PINALSIGYAHRPI" 解釋：P I NA L S I GY A H RP I我已經(jīng)編寫了以下代碼，但是在如何將行標(biāo)記為一次向下移動(dòng)方面我陷入了困境，我增加了起始行，但是當(dāng)它曲折回到頂部時(shí)，它應(yīng)該減少。我無(wú)法弄清楚在不影響向下運(yùn)動(dòng)的情況下進(jìn)行這項(xiàng)工作的邏輯。任何幫助，將不勝感激。const convert = (s, numRows) => { let startRow = 0 let endRow = numRows - 1 let startColumn = 0 let endColumn = Math.floor((s.length / 2) - 1) s = s.split('') let results = [] // to setup the columns for (let i = 0; i < numRows; i++) { results.push([]) } while (startRow <= endRow && startColumn <= endColumn && s.length) { for (let i = startRow; i <= endRow; i++) { results[i][startColumn] = s.shift() } for (let i = endRow - 1; i >= startRow; i--) { results[i][startColumn + 1] = s.shift() startColumn++ } //this line seems to be the issue startRow++ } return results}console.log(convert('PAYPALISHIRING', 4))

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2 回答

拉丁的傳說(shuō)

TA貢獻(xiàn)1789條經(jīng)驗(yàn) 獲得超8個(gè)贊

我重寫了您的 while 循環(huán)，如下所示，我只是走“之字形”模式！希望它足夠簡(jiǎn)單易懂。

let c=0, row=0,col=0, down=0;

while(c<s.length) {

results[row][col]=s[c];

if(down==0) { // moving down

row++;

if(row==numRows) {

down = 1;

col++;

row-=2;

}

} else { // moving up

row--;

col++;

if(row==0) {

down=0;

}

c++;

}

附言。上面的代碼無(wú)法處理numRows < 3，因此您必須在此循環(huán)之前管理它們。

反對(duì) 回復(fù) 2022-06-09

德瑪西亞99

TA貢獻(xiàn)1770條經(jīng)驗(yàn) 獲得超3個(gè)贊

我的微積分有點(diǎn)生疏，但這個(gè)問(wèn)題背后的邏輯似乎是一個(gè)正弦波。我在創(chuàng)建阻止它工作的 sin 方程的某個(gè)地方出現(xiàn)了數(shù)學(xué)錯(cuò)誤（r 從不等于 c 與當(dāng)前參數(shù)），但如果這是您選擇的方向，希望這會(huì)有所幫助。

/*If x-axis is position in string, and y-axis is row number...

n=number of rows

Equation for a sin curve: y = A sin(B(x + C)) + D

D=vertical shift (y value of mid point)

D=median of 1 and n

n: median:

1 1

2 1.5

3 2

4 2.5

5 3

6 3.5

7 4

median=(n+1)/2

D=(n+1)/2

A=amplitude (from the mid-point, how high does the curve go)

median + amplitude = number of rows

amplitude = number of rows - median

A=n-D

C=phase shift

Phase shift for a sin curve starting at its lowest point: 3π/2

(so at time 1, row number is 1, and curve goes up from there)

C=3π/2

Period is 2π/B

n p

3 4

4 6

5 8

6 10

period=2(n-1)

2(n-1)=2π/B

B(2(n-1)=2π

B=2π/2(n-1)

B=π/(n-1)

Variables:

s = string

n = number of rows

c = current row number being evaluated

p = position in string

r = row number

*/

var output='';

function convert(s,n) {

D=(n+1)/2

A=n-D

C=(3*Math.PI)/2

B=Math.PI/(n-1)

for (c=1;c<=n;c++) { //loop from 1st row to number of rows

for (p=1;p<=s.length;p++) { //loop from 1st to last character in string

r=A*Math.sin(B*(p+C))+D //calculate the row this character belongs in

if (r==c) { output+= s.charAt(r) } //if the character belongs in this row, add it to the output variable. (minus one because character number 1 is at position 0)

}}

//do something with output here

}

反對(duì) 回復(fù) 2022-06-09