标签:leetcode
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
.
解题报告:
用个变量last保存遍历到当前位置的最后一个空格位置,只要当前字符不是空格,就更新
res,这样下来,最终res里面存放的就是所要求的结果.需要注意的是,last变量的初值.
class Solution { public: int lengthOfLastWord(const char *s) { if (s == NULL) return 0; int res = 0,last = -1; for (int i = 0; s[i]; ++i) { if( s[i] == ' ') { last = i; continue; } res = i - last; } return res; } };
LeetCode:Length of Last Word,布布扣,bubuko.com
标签:leetcode
原文地址:http://blog.csdn.net/dream_you_to_life/article/details/34471409