用“ ”替代string中所有的空格

solution

如果有足够的空间，可以开辟另外一个char数组，当遇到空格时，将空格替换为"%20"然后存入数组。

char str[1000];

如果没有多余的空间，假设原str数组足够大，可以遍历str数组，求出空格的个数spaceCount，根据空格的个数算出需要增加的空间数：spaceCount*2。然后从后边遍历原数组，遇到空格时进行替换，并放入原数组的后边。

The algorithm is as follows:1. Count the number of spaces during the first scan of the string.2. Parse the string again from the end and for each character:»» If a space is encountered, store “%20”.
»» Else, store the character as it is in the newly shifted location.

假设数组在后边有足够的空间

// Assume parsed string has sufficient free space at the end

ReplaceFun(char * str, int length) {
int SpaceCount = 0, NewLength, i=0;
//计算空格数目
for (i = 0; i < length; i++) {
if (str[i] == ’ ‘) {
SpaceCount++;
}
}
//重新计算需要的空间数目
NewLength = length + SpaceCount * 2;
Str[NewLength] = ‘\0’;
//从后面开始写入新数据
for (i = length — 1; i >= 0; i--) {
if (str[i] == ‘ ‘) {
str[NewLength -1 ] = ‘0‘;
str[NewLength -2 ] = ‘2’;
str[NewLength -3 ] = ‘%’;
NewLength = NewLength -3;
} else {
str[NewLength - 1] = str[i];
NewLength = NewLength -1;
}
}
}