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ZigZag Conversion

时间:2014-09-07 21:07:35      阅读:247      评论:0      收藏:0      [点我收藏+]

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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".

思路:若为第一行或最后一行(即i=0 or i=nRow-1),则同一行的相邻元素索引相差2*(nRow-1);若为中间行(即0<i<nRow-1),则相邻索引相差为2*(nRow-1-i)和2i的交替。

 1 class Solution {
 2 public:
 3     string convert( string s, int nRows ) {
 4         if( --nRows <= 0 ) { return s; }
 5         int slen = s.length();
 6         string zigzag = "";
 7         for( int i = 0; i <= nRows; ++i ) {
 8             if( i == 0 || i == nRows ) {
 9                 for( int k = i; k < slen; k += 2*nRows ) {
10                     zigzag += s[k];
11                 }
12             } else {
13                 for( int k = i; k < slen; k += 2*nRows ) {
14                     zigzag += s[k];
15                     if( k+2*(nRows-i) < slen ) { zigzag += s[k+2*(nRows-i)]; }
16                 }
17             }
18         }
19         return zigzag;
20     }
21 };

 

ZigZag Conversion

标签:style   blog   color   io   for   div   sp   log   on   

原文地址:http://www.cnblogs.com/moderate-fish/p/3960724.html

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