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题目：//要求写一个函数，将字符串中的空格替换为%20.
样例：“abc efgx yz"转换成"abc%20efgx%20yz”

#include<stdio.h> #include<windows.h> #include<assert.h> void replace(char *src) { assert(src); int oldlen = 0; //原字符串长度 int newlen = 0; //新字符串长度 int blacknum = 0; //空格数量 int newback = 0; //新字符串尾部 int oldback = 0; //原字符串尾部 while (src[oldlen] != '\0') { if (src[oldlen] == ' ') { blacknum++; } oldlen++; } newlen = oldlen + blacknum * 2; oldback = oldlen - 1; newback = newlen - 1; while (oldback != 0){ if (src[oldback] == ' '){ src[newback--] = '0'; src[newback--] = '2'; src[newback] = '%'; } else{ src[newback] = src[oldback]; } oldback--; newback--; } } int main(){ char p[20] = "abc defgx yz"; replace(p); printf("%s\n", p); system("pause"); return 0; }

运行结果：