Code & Func

第76天。

快考试，可能要水一个月的easy题了。

Given a string s consists of upper/lower-case alphabets and empty space characters ’ ’, return the length of last word in the string.

If the last word does not exist, return 0.

Note: A word is defined as a character sequence consists of non-space characters only.

Example:

Input: “Hello World” Output: 5

一看完题目，我就想到了python的split:

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def lengthOfLastWord(self, s):
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    """
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    :type s: str
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    :rtype: int
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    """
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    words = s.split();
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    if len(words) == 0: return 0;
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    return len(words[-1])

然后是用c++用find去解的：

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int lengthOfLastWord(string s) {
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    auto beg = s.begin();
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    auto it = beg;
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    auto end = s.end();
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    // fix the bug like that "hello world "
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    for(int i = s.size() - 1;i >= 0 && s[i] == ' ';i--)
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        end--;
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    while((it = find(beg,end,' ')) != end) {
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        beg = it + 1;
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    }
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    return end - beg;
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}

然后是从后面向前扫描的方法：

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int lengthOfLastWord(string s) {
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    auto end = s.rbegin();
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    while(end != s.rend() && *end == ' ') end++;
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    auto beg = end;
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    while(beg != s.rend() && *beg != ' ') beg++;
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    return beg - end;
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}

然后是dicuss中的解法，和上面的从后向前扫描的方法类似，只不过它第二个循环里面顺带计算了length:

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int lengthOfLastWord(string s) {
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    int len = 0, tail = s.length() - 1;
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    while (tail >= 0 && s[tail] == ' ') tail--;
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    while (tail >= 0 && s[tail] != ' ') {
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        len++;
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        tail--;
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    }
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    return len;
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}