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6. Zigzag Conversion


6. Zigzag Conversionopen in new window

🟠   🔖  字符串  🔗 LeetCodeopen in new window

题目

The string "PAYPALISHIRING" is written in a zigzag pattern on a given number of rows like this: (you may want to display this pattern in a fixed font for better legibility)

P   A   H   N
A P L S I I G
Y   I   R

And then read line by line: "PAHNAPLSIIGYIR"

Write the code that will take a string and make this conversion given a number of rows:

string convert(string s, int numRows);

Example 1:

Input: s = "PAYPALISHIRING", numRows = 3

Output: "PAHNAPLSIIGYIR"

Example 2:

Input: s = "PAYPALISHIRING", numRows = 4

Output: "PINALSIGYAHRPI"

Explanation:

P     I    N
A   L S  I G
Y A   H R
P     I

Example 3:

Input: s = "A", numRows = 1

Output: "A"

Constraints:

  • 1 <= s.length <= 1000
  • s consists of English letters (lower-case and upper-case), ',' and '.'.
  • 1 <= numRows <= 1000

题目大意

将一个给定字符串 s 根据给定的行数 numRows ,以从上往下、从左到右进行  Z 字形排列。

比如输入字符串为 "PAYPALISHIRING"  行数为 3 时,排列如下:

P   A   H   N
A P L S I I G
Y   I   R

之后,你的输出需要从左往右逐行读取,产生出一个新的字符串,比如:"PAHNAPLSIIGYIR"

请你实现这个将字符串进行指定行数变换的函数:

string convert(string s, int numRows);

解题思路

  • 这一题没有什么算法思想,考察的是对程序控制的能力;
  • 用 2 个变量保存方向,当垂直输出的行数达到了规定的目标行数以后,需要从下往上转折到第一行;
  • 循环中控制好方向,注意转变方向时的情况处理。

代码

/**
 * @param {string} s
 * @param {number} numRows
 * @return {string}
 */
var convert = function (s, numRows) {
  let arr = new Array(numRows).fill("");
  // 负责从第一行开始向下加,直到第numRows行
  let down = 0;
  // 负责从倒数第一行开始向上加,直到第二行
  let up = numRows - 2;
  for (let i = 0; i < s.length; i++) {
    if (down < numRows) {
      arr[down] += s[i];
      down++;
    } else if (up > 0) {
      arr[up] += s[i];
      up--;
    } else {
      arr[0] += s[i];
      down = 1;
      up = numRows - 2;
    }
  }
  return arr.join("");
};