[LeetCode] 58. Length of Last Word 求末尾单词的长度
2018-09-07 04:39
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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.
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; } };
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