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替换空格问题_萌新的日常的技术博客_51CTO博客
source link: https://blog.51cto.com/u_15787387/5780340
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替换空格问题
精选 原创实现一个函数，将一个字符串的每个空格替换成"%20",例如，当字符串为 We Are Happy.则经过替换之后的字符串为 We%20Are%20Happy.
>#include<stdio.h>
#include<string.h>
void replaceSpace(char* str, int length)
{
int sum = 0;
char* cur = str;
while (*cur != '\0')
{
if (*cur == ' ')
{
sum++;
}
cur++;
}
char* end1 = str + length - 1;
char* end2 = str + length + 2 * sum - 1;
while (end1 != end2)
{
if (*end1 != ' ')
{
*end2-- = *end1--;
}
else
{
*end2-- = '0';
*end2-- = '2';
*end2-- = '%';
end1--;
}
}
}
int main()
{
char arr[40] = "We Are Happy";
int len = strlen(arr);
replaceSpace(arr, len);
printf("%s\n", arr);
return 0;
}
刚开始时，根据空格的个数，来决定向后要走几个2
如图，若有两个空格，则向后移4个单位
当end1指向空格时，此时需要将 % 2 0 分别逆置 即打印 0 2 %在end2所指向处，
如图一，刚开始end1处于空格处，end2处于p处，
当将三个字符打印完毕后， end1–，end2–，如图二，end1处于e处，end2处于H处
此时end1处于e处，因为不是空格，所以将所在的数据传递给end2处
如图一，当end1再次遇见空格时，end2处于r处，完成对于 % 2 0三个字符的逆置放入end2中
如图二，在完成逆置 后，end1-- ，end2-- ，此时end1与end2同时处于e处
两者处于同一个位置，将同时完成
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