标签:des style color os strong io for ar
Problem Description:
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"
.分析:刚开始没看懂题意,以为是按照间隔多少每行输出,在那数啊数发现有点不对劲,在网上搜索了一下才发现是按照“之”字输出,具体如下图:
1 7 .
2 6 8 .
3 5 9 .
4 10
题目其实就是一个找规律的题目,从图中可以看出,假设指定输出行数为n,第一行和最后一行其实每个周期的元素之间相差2n-2个间隔,中间每行在两个间隔之间多了一个数,这个数也是有规律的,与下个一个周期的开始一个数相差i个间隔,比如6=(2*4-2)-1,按照这个规律就可以把每行的字符直接打印出来了,具体实现代码如下:
class Solution { public: string convert(string s, int nRows) { string res,temp; int n=s.size(); if(n==0) return res; if(nRows<=1) return s; int i=0;//添加第一行 while(i<n) { res+=s[i]; i+=2*nRows-2; } for(i=1;i<nRows-1;i++)//添加中间行 { int j=i,r=2*nRows-2; while(j<n) { res+=s[j]; if(r-i<n) res+=s[r-i];//添加中间的数 j+=2*nRows-2; r+=2*nRows-2; } } i=nRows-1;//添加最后一行 while(i<n) { res+=s[i]; i+=2*nRows-2; } return res; } };
Leetcode--ZigZag Conversion,布布扣,bubuko.com
标签:des style color os strong io for ar
原文地址:http://blog.csdn.net/longhopefor/article/details/38322897