The String ZigZag Conversion Algorithms

  • 时间：2020-09-12 10:17:13
  • 分类：网络文摘
  • 阅读：79 次

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

The String ZigZag Conversion Algorithms

We can simulate the process by walking down first, then when it reaches the last row, bounces back in the reverse direction. We need arrays of strings as we walk, and append the characters to the corresponding row of string.

O(N) time and O(N) space requirement. Edge case is when the ZigZag row is one and then we have to return the original string.

C++ String ZigZag Conversion
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21

class Solution {
public:
    string convert(string s, int numRows) {
        if (numRows == 1) return s;
        vector<string> rows(min((int)s.size(), numRows));
        bool down = false;        
        int row = 0;
        for (int i = 0; i < s.size(); ++ i) {
            rows[row] += s[i];
            if ((row == 0) || (row == numRows - 1)) {
                down = !down;
            }
            row += down ? 1 : -1;
        }        
        string ss = "";
        for (const auto &n: rows) {
            ss += n;
        }
        return ss;
    }
};
class Solution {
public:
    string convert(string s, int numRows) {
        if (numRows == 1) return s;
        vector<string> rows(min((int)s.size(), numRows));
        bool down = false;        
        int row = 0;
        for (int i = 0; i < s.size(); ++ i) {
            rows[row] += s[i];
            if ((row == 0) || (row == numRows - 1)) {
                down = !down;
            }
            row += down ? 1 : -1;
        }        
        string ss = "";
        for (const auto &n: rows) {
            ss += n;
        }
        return ss;
    }
};
Python String ZigZag Conversion
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17

class Solution:
    def convert(self, s: str, numRows: int) -> str:
        if numRows == 1:
            return s
        data = [''] * numRows
        down = False
        row = 0
        for i in s:
            data[row] += i
            if row == 0 or row == numRows - 1:
                down = not down
            row = row + 1 if down else row - 1
        x = ''
        for j in data:
            if j != '':
                x = x + j
        return x
class Solution:
    def convert(self, s: str, numRows: int) -> str:
        if numRows == 1:
            return s
        data = [''] * numRows
        down = False
        row = 0
        for i in s:
            data[row] += i
            if row == 0 or row == numRows - 1:
                down = not down
            row = row + 1 if down else row - 1
        x = ''
        for j in data:
            if j != '':
                x = x + j
        return x
Javascript String ZigZag Conversion
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27

/**
 * @param {string} s
 * @param {number} numRows
 * @return {string}
 */
var convert = function(s, numRows) {
    if (numRows === 1) return s;
    let data = [];
    let row = 0;
    let down = false;
    for (let x of s) {
        if (typeof data[row] === 'undefined') {
            data[row] = x;
        } else {
            data[row] += x;
        }
        if (row === 0 || row === numRows - 1) {
            down = !down;
        }
        row += down ? 1 : -1;
    }
    let res = '';
    for (let i = 0; i < Math.min(numRows, s.length); ++ i) {
        res += data[i];
    }
    return res;
};
/**
 * @param {string} s
 * @param {number} numRows
 * @return {string}
 */
var convert = function(s, numRows) {
    if (numRows === 1) return s;
    let data = [];
    let row = 0;
    let down = false;
    for (let x of s) {
        if (typeof data[row] === 'undefined') {
            data[row] = x;
        } else {
            data[row] += x;
        }
        if (row === 0 || row === numRows - 1) {
            down = !down;
        }
        row += down ? 1 : -1;
    }
    let res = '';
    for (let i = 0; i < Math.min(numRows, s.length); ++ i) {
        res += data[i];
    }
    return res;
};

–EOF (The Ultimate Computing & Technology Blog) —

推荐阅读：
萤火之光，可以照明　 看得破的是人生作文200字　 学《弟子规》有感　 描写五莲山风景的作文　 诗词名句鉴赏：少壮不努力，老大徒伤悲。　 诗词名句鉴赏：秋风萧萧愁杀人　 诗词名句鉴赏：树欲静而风不止　 诗词名句鉴赏：忧艰常早至，欢会常苦晚。　 春王正月原文及翻译　 诗词名句鉴赏：风萧萧兮易水寒，壮士一去兮不复还!　
评论列表
添加评论