从就地一字符串中去除空白？ [英] Strip whitespace from a string in-place?

问题描述

我在的采访问题清单看到这个。让我疑惑。

I saw this in a "list of interview questions". Got me wondering.

不限于空白字符必须地，当然，很容易推广到除去从字符串，就地某些特定的字符。

Not limited to whitespace necessarily, of course, easily generalized to "removing some specific character from a string, in-place".

我的解决办法是：

void stripChar(char *str, char c = ' ') {
  int x = 0;
  for (int i=0;i<strlen(str);i++) {
    str[i-x]=str[i];
    if (str[i]==c) x++;
  }
  str[strlen(str)-x] = '\0';
}

我怀疑有一个更有效的，但有一个解决方案，更优雅？

I doubt there is one more efficient, but is there a solution that is more elegant?

修改：完全忘了我离开的strlen 在那里，这是最绝对不是有效的。

edit: totally forgot I left strlen in there, it is most definitely not efficient

推荐答案

首先， I＆LT;的strlen（STR）始终是遍历字符串低效的成语。正确的循环条件简直是海峡[I] ，即循环，直到海峡[I] 是空终止。

First of all, i<strlen(str) is always an inefficient idiom for looping over a string. The correct loop condition is simply str[i], i.e. loop until str[i] is the null terminator.

随着中说，这里是最简单/最简洁的算法，我才知道：

With that said, here's the simplest/most concise algorithm I know:

for (size_t i=0, j=0; s[j]=s[i]; j+=!isspace(s[i++]));

注意：我的解决方案是相对于身体（特定字符）写在主题（空格）的问题。如果需要，可以轻松适应它。

Note: My solution is for the question as written in the subject (whitespace) as opposed to the body (particular character). You can easily adapt it if needed.

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