首页 > itarticle > zigzagconversion

zigzagconversion

admin 11月 12, 2020 0 

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
 

 * 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". 
 */

直接模拟

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
 
class Solution {  
public:  
    string (string s, int nRows) {  
        if (nRows == 1) 
            return s;  
        //step = (nRows - 1) * 2
        int step = nRows * 2 - 2, len = s.length();  
        string ret = "";  
        //第一行  
        for (int i = 0; i < len; i += step)  
            ret += s[i];  
        //中间行
        for (int i = 1; i < nRows - 1; i++) {  
            for (int j = i; j < len; j += step) {  
                ret += s[j];  
                if (j + (step - i * 2) < len)  
                    ret += s[j + (step - i * 2)];  
            }  
        }  
        //最后一行 
        for (int i = nRows - 1; i < len; i += step)  
            ret += s[i];  
        return ret;  
    }  
};
 
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
 
public class ZigZagConversion {
    public String (String s, int nRows) {
        if (nRows == 1)
            return s;
        String res = "";
        int len = s.length();
        int T = 2 * nRows - 2;
        for (int i = 0; i < nRows; i++) {
            for (int j = 0; i + j * T < len; j++) {
                res += s.charAt(i + j * T);
                if (i > 0 && i < nRows - 1 && (T - i + j * T < len))
                    res += s.charAt(T - i + j * T);
            }
        }
        return res;
    }
}
微海报分享

近期文章

近期评论

标签

热门

文章归档

分类目录

功能