[LeetCode] 58. Length of Last Word 求末尾单词的长度

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

Java:

public int lengthOfLastWord(String s) {
    String use = s.trim();
    int count = 0;
    for (int i = use.length() - 1; i >= 0; i--) {
        if (use.charAt(i) != ‘ ‘) count++;
        else break;
    }
    return count;
}　　

Java:

public int lengthOfLastWord(String s) {
    return s.trim().length()-s.trim().lastIndexOf(" ")-1;
}　　

Python:

class Solution(object):
    def lengthOfLastWord(self, s):
        """
        :type s: str
        :rtype: int
        """
        return len(s.strip().split(‘ ‘)[-1])

Python:

class Solution:
    # @param s, a string
    # @return an integer
    def lengthOfLastWord(self, s):
        length = 0
        for i in reversed(s):
            if i == ‘ ‘:
                if length:
                    break
            else:
                length += 1
        return length

C++:

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

原文地址：https://www.cnblogs.com/lightwindy/p/9602354.html