标签：style blog class c code tar
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 RAnd 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行,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：
|
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22 |
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; }}; |
【版权声明】转载请注明出处：http://www.cnblogs.com/TenosDoIt/p/3738693.html
LeetCode:ZigZag Conversion,布布扣,bubuko.com
标签：style blog class c code tar
原文地址：http://www.cnblogs.com/TenosDoIt/p/3738693.html
