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".| 0 | 8 | 16 | |||||||||
| 1 | 7 | 9 | 15 | 17 | |||||||
| 2 | 6 | 10 | 14 | 18 | |||||||
| 3 | 5 | 11 | 13 | 19 | |||||||
| 4 | 12 | 20 |
|
| 0 | 8 | 16 | |||||||||
| 1 | 7 | 9 | 15 | 17 | |||||||
| 2 | 6 | 10 | 14 | 18 | |||||||
| 3 | 5 | 11 | 13 | 19 | |||||||
| 4 | 12 | 20 |
|
public class Solution {
public String convert(String s, int numRows) {
if(numRows==1)return s;
char[] chs=s.toCharArray();
if(numRows==2){
int index=0;
for(int i=0;i<s.length();i=i+2)chs[index++]=s.charAt(i);
for(int i=1;i<s.length();i=i+2)chs[index++]=s.charAt(i);
return new String(chs);
}
char[][] map = new char[numRows][s.length()];
int index=0,col=0;
while(index<s.length()){
if(col%2==1){
int i=numRows-2;
while(i>=1 && index<s.length())map[i--][col]=s.charAt(index++);
}else{
int i=0;
while(i<numRows && index<s.length())map[i++][col]=s.charAt(index++);
}
col++;
}
index=0;
for(int i=0;i<numRows;i++){
for(int j=0;j<col;j++){
if(map[i][j]!=0)chs[index++]=map[i][j];
}
}
return new String(chs);
}
}C语言源代码（用时79ms）：char* convert(char* s, int numRows) {
char* ch;
char map[1000][1000];
int length=0,index=0,i,j,k;
if(numRows<=1)return s;
for(i=0;s[i];i++)length++;
ch = (char*)malloc(sizeof(char)*length*2);
if(numRows==2){
for(i=0;i<length;i=i+2)ch[index++]=s[i];
for(i=1;i<length;i=i+2)ch[index++]=s[i];
ch[index]=0;
return ch;
}
for(i=0;i<numRows;i++){
for(j=0;j<length;j++){
map[i][j]=0;
}
}
i=0,j=0,index=0;
while(s[index]){
if(j%2){
i=numRows-2;
while(i>=1 && s[index])map[i--][j]=s[index++];
}else{
i=0;
while(i<numRows && s[index])map[i++][j]=s[index++];
}
j++;
}
index=0;
for(i=0;i<numRows;i++){
for(k=0;k<j;k++){
if(map[i][k]!=0)ch[index++]=map[i][k];
}
}
ch[index]=0;
return ch;
}C++源代码（用时112ms）：class Solution {
public:
string convert(string s, int numRows) {
if(numRows==1)return s;
string chs=string(s);
int index=0;
if(numRows==2){
for(int i=0;i<s.size();i=i+2)chs[index++]=s[i];
for(int i=1;i<s.size();i=i+2)chs[index++]=s[i];
return chs;
}
char map[1000][1000];
for(int i=0;i<numRows;i++){
for(int j=0;j<s.size();j++)
map[i][j]=0;
}
index=0;
int col=0;
while(s[index]){
if(col%2){
int i=numRows-2;
while(i>=1 && s[index])map[i--][col]=s[index++];
}else{
int i=0;
while(i<numRows && s[index])map[i++][col]=s[index++];
}
col++;
}
index=0;
for(int i=0;i<numRows;i++){
for(int j=0;j<col;j++){
if(map[i][j])chs[index++]=map[i][j];
}
}
return chs;
}
};Python源代码（用时138ms）：class Solution:
# @param {string} s
# @param {integer} numRows
# @return {string}
def convert(self, s, numRows):
if numRows==1:return s
res=["" for i in range(numRows)]
i=0;gap=numRows-2
while i<len(s):
j=0
while i<len(s) and j<numRows:res[j]+=s[i];i+=1;j+=1
j=gap
while i<len(s) and j>0:res[j]+=s[i];i+=1;j-=1
chs=''
for i in range(numRows):
chs+=res[i]
return chsLeetCode 6 ZigZag Conversion (C,C++,Java,Python)
原文地址：http://blog.csdn.net/runningtortoises/article/details/45542565
