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”.
int main() { string s; int nrow; while (cin >> s >> nrow) { char imap[1000][1000]; //数组大小尝试出来的 memset(imap, 0, sizeof(imap)); int pos = 0; int i = 0; while (pos < s.size()) { if (i % 2 == 0) //列为偶数时 { for (int k = 0; k < nrow && s[pos]!='\0'; k++) //注意添加的时候，要判断s[pos]!='\0' ,如果等于的话，则也跳出了循环，因为pos此时等于s.size()了 { imap[k][i] = s[pos++]; } } else //列为奇数时 { if(s[pos] != '\0') //判断s[pos]!='\0' { if (nrow == 1) //nrow == 1另外考虑 ABCD 1 --> ABCD imap[0][i] = s[pos++]; else if (nrow == 2) //nrow == 2另外考虑 ABCD 2 -->ACBD for (int k = 0; k < nrow && s[pos] != '\0'; k++) { imap[k][i] = s[pos++]; } else //nrow > 2, 从nrow - 2由下往上排 for (int k = nrow - 2; k >= 1 && s[pos] != '\0'; k--) { imap[k][i] = s[pos++]; } } } i++; } s.clear(); for (int r = 0; r < nrow; r++) for (int c = 0; c < i; c++) { if (imap[r][c] != 0)s.push_back(imap[r][c]); } cout << s << endl; } }