題目
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".
就是將一個(gè)字符串Z字形排列后再讀出的字符串穷遂。
可以依次得到各個(gè)行的字符串再拼接在岂，比較簡(jiǎn)單例证。
我是直接計(jì)算的字符串的每位是怎么映射到結(jié)果字符串的。都可以乓梨。
下面的C代碼，已通過(guò)。
char* convert(char* s, int numRows) {
int length=0,i=0,k=0,ansk=0;
while(s[length]!='\0')length++;
if(numRows==1||numRows>=length)return s;
char * ans=(char *)malloc((length+1)*sizeof(char));
for(i=0;i<numRows;i++)
{
k=i;
if(i==0||i==numRows-1)
{
while(length>k&&length>ansk)
{
ans[ansk]=s[k];k+=2*(numRows-1);ansk++;
}
}
else
{
while(length>k&&length>ansk)
{
ans[ansk]=s[k];ansk++;
k+=2*(numRows-i-1);
if(k>=length||ansk>=length)break;
ans[ansk]=s[k];ansk++;
k+=2*(i);
}
}
// printf("%s\n",ans);
}
ans[length]='\0';
return ans;
}
