0006. ZigZag Conversion (M)

ZigZag Conversion (M)

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

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题意

给定字符串，将其按照指定行数进行之字形的重新排列，按顺序输出新图形的每一行字符。

思路

直接模拟字符重排的过程，按顺序将每一个字符存入对应行号的字符串中，注意在第一行和最后一行处要改变行号变化的方向。

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代码实现

Java

class Solution {
    public String convert(String s, int numRows) {
        // 特殊情况直接返回字符串
        if (numRows == 1) {
            return s;
        }
        
        // 分别保存重排后的每一行
        List<StringBuilder> sbs = new ArrayList<>();
        for (int i = 0; i < numRows; i++) {
            sbs.add(new StringBuilder());
        }
        
        // row指定当前字符对应的行号
        int row = 0;
        // flag用于判断行号增减方向
        boolean flag = false;
        // 按顺序将字符保存到对应的行字符串中
        for (char c : s.toCharArray()) {
            sbs.get(row).append(c);
            // 在第一行和最后一行要变换行号增减方向
            if (row == 0 || row == numRows - 1) {
                flag = !flag;
            }
            if (flag) {
                row++;
            } else {
                row--;
            }
        }
        
        StringBuilder res = new StringBuilder();
        for (StringBuilder sb : sbs) {
            res.append(sb);
        }
        return res.toString();
    }
}

JavaScript

/**
 * @param {string} s
 * @param {number} numRows
 * @return {string}
 */
var convert = function (s, numRows) {
  if (numRows === 1) {
    return s
  }

  let rows = []
  for (let i = 0; i < numRows; i++) {
    rows[i] = []
  }
  let line = 0
  let flag = false
  for (let c of s) {
    rows[line].push(c)
    if (line === 0 || line === numRows - 1) {
      flag = !flag
    }
    line += flag ? 1 : -1
  }
  return rows.map((chars) => chars.join('')).join('')
}