本文主要是介绍字形变换(字符串)，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
题目来自力扣：
做之前我必须吐槽几句，这是我写过最难的题了！！！
下面是我的代码
别问为什么没有注释，我打代码从来不带注释，看得懂的才是同路人
char* convert(char* s, int numRows)
{int sLenth = strlen(s);printf("字符串的长度为：%d\n", sLenth);char* output = (char*)malloc(1000);memset(output, 0, 1000);int subscript = 0;int index = sLenth / (numRows + (numRows - 2));index++;for (int i = 0; i < index; i++){output[i] = s[subscript];subscript += numRows + (numRows - 2);}subscript = 0;int num = 0;int arrayLength = strlen(output);int index2 = 1;//numRowsfor (int i = 0; i < numRows - 1; i++){for (int j = 0; j < (index + (index - 1)) / 2 + 1; j++){if (s[num + index2] != '\0'){printf("s[%d + %d] == %c\n", num, index2, s[num + index2]);output[arrayLength++] = s[num + index2];}if (s[num + numRows + numRows - 2 - 1] != '\0'){printf("s[%d+%d+%d-2-%d] == %c\n",num,numRows,numRows,index2, s[num + numRows + numRows - 2 - index2]);if ((num + index2) == (num + numRows + numRows - 2 - index2)){goto A;}output[arrayLength++] = s[num + numRows + numRows - 2 - index2];} A:;printf("\n");num += (numRows + numRows - 2);}num = 0;index2++;}return output;
}int main()
{int numRows = 0;char* s = (char*)malloc(1000);int sLenth = strlen(s);memset(s, 0, 1000);printf("请用户输入字符串：");scanf("%s", s);while (getchar() != '\n');printf("请用户输入行数：");scanf("%d", &numRows);if (numRows == 1){printf("字形变化后的字符串为：");printf("%s", s);return 0;}char* op = convert(s, numRows);printf("字形变化后的字符串为：");for (int i = 0; i < sLenth; i++){printf("%c", op[i]);}return 0;
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