题目:
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:
假设有ABCDEFG这样的字符串,我们可以表示为如下格式。即分为上行和下行两列,考虑上行下行两种情况以及对应的游标跳转的表达式。
A G
B F
C E
D
代码:
方法1:
class Solution {
public:
string convert(string s, int nRows) {
if(nRows==1)
{
return s;
}
else
{
string str;
for(int i=0;i<nRows;i++)
{
str += convertLine(i, nRows, s, i!=(nRows-1));
}
return str;
}
}
string convertLine(int num, int nRows, string s, bool downstair)
{
int cur = num;
string str;
while(cur<s.size())
{
str.push_back(s[cur]);
if(downstair)
{
cur+= 2*(nRows-1-cur%(nRows-1));
}
else
{
cur+=2*(nRows-1-cur%(nRows-1));
}
};
return str;
}
};
方法2:
class Solution {
public:
string result;
string convert(string s, int nRows) {
if(nRows <= 1){
return s;
}
int unit = 2 * nRows - 2;
int n = s.size() / unit;
for(int i = 0; i < nRows; i++){
for(int j = 0; j <= n; j++){
if(j * unit + i < s.size() && (i == 0 || i == nRows - 1))
result.push_back(s[j * unit + i]);
else{
if(j * unit + i < s.size())
result.push_back(s[j * unit + i]);
if(j * unit + unit - i < s.size())
result.push_back(s[j * unit + unit - i]);
}
}
}
return result;
}
};