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)this
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:code
string convert(string text, int nRows);
convert("PAYPALISHIRING", 3) should return "PAHNAPLSIIGYIR". input
zigzag的排列规律:第一行和最后一行都是没有斜边的,间隔是(2*nRows - 2)其中nRows是总行数;string
有斜边的中间几行其实也有规律,斜边上的数字到左边那个列的距离是(2nRows - 2 - 2row)其中row是当前行的行号。it
#include <stdio.h> #include <stdlib.h> #include <string.h> char *convert(char *s, int nRows) { if ((NULL == s) | (nRows < 1)) { return NULL; } // + 1 for NIL or '\0' in the end of a string const size_t len = strlen(s); char* output = (char*) malloc(sizeof(char) * ( len + 1)); char* head = output; output[len] = '\0'; if ( 1 == nRows ) { return strcpy(output, s); } for (int row = 0; row < nRows; ++row) { //processing row by row using (2nRows-2) rule for (unsigned int index = row; index < len; index += 2*nRows-2) { // if it is the first row or the last row, then this is all *output++ = s[index]; // otherwise, there are middle values, using (2nRows-2-2*row) rule // notice that nRows-1 is the last row if ( (row>0)&(row<nRows-1) & ((index+2*nRows - 2 - 2*row) < len)) { *output++ = s[index+2*nRows - 2 - 2*row]; } } } return head; } int main() { char* input = (char*)"A"; int rows = 1; char* output = convert(input, rows); if (NULL != output) { printf("input: %s; output: %s\n", input, output); free(output); }else { printf("empty\n"); } }