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".
这道题就是看坐标的变化，找规律并分块处理。参考：爱做饭的小莹子
规律：第一行和最后一行，就是按照2n-2的顺序一点点加的。斜着那条线的字的位置是当前列j+(2n-2)-2i(i是行的index）。
Java：
public String convert(String s, int nRows) {
if(s == null || s.length()==0 || nRows <=0)
return "";
if(nRows == 1)
return s;
StringBuilder res = new StringBuilder();
int size = 2*nRows-2;
for(int i=0;i<nRows;i++){
for(int j=i;j<s.length();j+=size){
res.append(s.charAt(j));
if(i != 0 && i != nRows - 1){//except the first row and the last row
int temp = j+size-2*i;
if(temp<s.length())
res.append(s.charAt(temp));
}
}
}
return res.toString();
}
Python:
class Solution(object):
def convert(self, s, numRows):
"""
:type s: str
:type numRows: int
:rtype: str
"""
if numRows == 1:
return s
step, zigzag = 2 * numRows - 2, ""
for i in xrange(numRows):
for j in xrange(i, len(s), step):
zigzag += s[j]
if 0 < i < numRows - 1 and j + step - 2 * i < len(s):
zigzag += s[j + step - 2 * i]
return zigzag
C++:
class Solution {
public:
string convert(string s, int nRows) {
if (nRows <= 1) return s;
string res = "";
int size = 2 * nRows - 2;
for (int i = 0; i < nRows; ++i) {
for (int j = i; j < s.size(); j += size) {
res += s[j];
int tmp = j + size - 2 * i;
if (i != 0 && i != nRows - 1 && tmp < s.size()) res += s[tmp];
}
}
return res;
}
};
