【LeetCode】6. ZigZag Conversion

最新推荐文章于 2021-03-20 20:07:40 发布

原创最新推荐文章于 2021-03-20 20:07:40 发布 · 323 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#leetcode

leetcode 专栏收录该内容

39 篇文章

订阅专栏

本文介绍了一种将字符串以ZigZag形式分布到多行再按行读取的算法实现。通过示例“PAYPALISHIRING”展示了如何在指定行数下进行转换，并提供了C++代码实现。

6. ZigZag Conversion

My Submissions

Total Accepted: 129677
Total Submissions: 500836
Difficulty: Easy
Contributors: Admin

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 text, int nRows);

convert("PAYPALISHIRING", 3) should return "PAHNAPLSIIGYIR".

Subscribe to see which companies asked this question

向下循环:nRows
斜角线循环:nRows-2(减去首尾两个端点)
重复

string convert(string s, int nRows){
    if(nRows == 1) return s;
    string res[nRows];
    int i = 0, j, gap = nRows-2;
    while(i < s.size()){
        for(j = 0; i < s.size() && j < nRows; ++j) res[j] += s[i++];
        for(j = gap; i < s.size() && j > 0; --j) res[j] += s[i++];
    }
    string str = "";
    for(i = 0; i < nRows; ++i)
        str += res[i];
    return str;
}