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  • [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;
        }
    };
    

     

     

    All LeetCode Questions List 题目汇总

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