LeetCode::Zigzag Conversion C语言

来源：互联网发布：js 滑动特效编辑：程序博客网时间：2024/05/15 23:24

题目

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   NA P L S I I GY   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".

分析

其实只有一个难点，只有发现了规律就很简单了。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；

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（2*nRows - 2 - 2*row）其中row是当前行的行号。

明白这两个规律就行了。

C语言代码注意就算s只有一行，也不能return s；要malloc 内存 strcpy 再返回。

#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");

0 0