leetcode解题: Length of Last Word (58)

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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.

For example,
Given s = “Hello World”,
return 5.

解法1：

首这题的启发，也用istringstream来读取一个个的word，直到最后一个string。
C++
```
这里要注意的是， getline会读取word直到下一个delimiter的出现，那么如果有两个连续的space，第二个space之前会读出一个“”， 也就是空字符串。

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class Solution {
public:
    int lengthOfLastWord(string s) {
        istringstream ss(s);
        string word;
        int res = 0;
        while (getline(ss, word, ' ')) {
                if (!word.empty()) {
                    res = word.size();    
                }
        }
        return res;
    }
};

Java

1

解法2：

不用istringstream的操作， 一位位的读。
C++

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class Solution {
public:
    int lengthOfLastWord(string s) {
        int res = 0;
        int count = 0;
        for (int i = 0; i < s.size(); ++i) {
            if (s[i] == ' ') {
                if (count != 0) {
                    res = count;
                } 
                count = 0;
            } else {
                ++count;
            }
        }
        if (count) res = count;
        return res;
    }
};