替換字串的空格

將字串中的空格替換成“123”。

顯然不能從前往後替換，這樣的時間複雜度是n2的。有兩種方法：

#include <stdio.h>#include <stdlib.h>#include <string.h>//如果原字串記憶體區不夠長，就重新開闢記憶體char* replace1(char *source,char *dest){int count = 0;char *temp1 = source,*temp2;if(source == NULL)return 0;for(;*temp1 != '\0';temp1++)if(*temp1 == ' ')count++;dest = malloc((strlen(source) + 2 * count + 1) * sizeof(char));temp1 = source;temp2 = dest;for(;*temp1 != '\0';temp1++,temp2++){if(*temp1 == ' '){*temp2++ = '1';*temp2++ = '2';*temp2 = '3';}else*temp2 = *temp1;}*temp2 = '\0';return dest;}//如果原字串記憶體區夠長，就在原記憶體處修改char* replace2(char *source){int count = 0,front,end;char *temp = source;if(source == NULL)return 0;for(;*temp != '\0';temp++)if(*temp == ' ')count++;front = strlen(source);end = strlen(source) + count * 2;for(;front >= 0 && end > front;front--,end--){if(source[front] == ' '){source[end--] = '3';source[end--] = '2';source[end] = '1';}elsesource[end] = source[front];}}void main(){char *source = "abc d ef",*dest;char s[20] = "abc d ef";dest = replace1(source,dest);printf("%s\n",source);printf("%s\n\n",dest);replace2(s);printf("%s\n",source);printf("%s\n",s);}