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".
题目的意思是把字符串上下上下走之字形状,然后按行输出,比如包含数字0~22的字符串, 给定行数为5,走之字形如下:
现在要按行输出字符,即:0 8 16 1 7 9 15 17 2…….
如果把以上的数字字符看做是字符在原数组的下标, 给定行数为n = 5, 可以发现以下规律:
(1)第一行和最后一行下标间隔都是interval = n*2-2 = 8 ; 本文地址
(2)中间行的间隔是周期性的,第i行的间隔是: interval–2*i, 2*i, interval–2*i, 2*i, interval–2*i, 2*i, …
代码如下，时间复杂度为O（n），n是字符串的长度：
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class
Solution {
public
:
string convert(string s,
int
nRows) {
if
(nRows == 1)
return
s;
int
len = s.size(), k = 0, interval = (nRows<<1)-2;
string res(len,
' '
);
for
(
int
j = 0; j < len ; j += interval)
//处理第一行
res[k++] = s[j];
for
(
int
i = 1; i < nRows-1; i++)
//处理中间行
{
int
inter = (i<<1);
for
(
int
j = i; j < len; j += inter)
{
res[k++] = s[j];
inter = interval - inter;
}
}
for
(
int
j = nRows-1; j < len ; j += interval)
//处理最后一行
res[k++] = s[j];
return
res;
}
};
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本文转自tenos博客园博客，原文链接：http://www.cnblogs.com/TenosDoIt/p/3738693.html，如需转载请自行联系原作者
