LeetCode58：Length of Last Word

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

---

LeetCode：链接

从后往前遍历。首先找到字符串中第一不为’ ‘的值的坐标，该坐标即为最后一个Word的尾字母。然后继续遍历，如果碰到’ ‘，则遍历结束，否则，count++。最后返回count。

class Solution:
    def lengthOfLastWord(self, s):
        """
        :type s: str
        :rtype: int
        """
        count = 0
        end = len(s) - 1
        '''从后往前遍历 直到找到不为空的字符'''
        while end >= 0 and s[end]==' ':
            end -= 1
        '''继续从后往前遍历 直到找到下一个不为空的字符'''
        while end >= 0 and s[end]!=' ':
            end -= 1
            count += 1
        return count