题目:
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
.
代码:
class Solution { public: int lengthOfLastWord(string s) { for ( int i = s.length()-1; i >=0; --i ) { if (s[i]==' ') { s.erase(s.end()-1); } else { break; } } const size_t len = s.length(); int ret = 0; for ( size_t i = 0; i < len; ++i ) { if (s[i]!=' ') { ++ret; continue; } if (s[i]==' ') { ret = 0; continue; } } return ret; } };
tips:
先把后面的空格都去掉。然后从头遍历,最后留下的ret就是最后一个单词的长度。
还有STL的一个做法,明天再看。
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第二次过这道题,感觉不知道第一次为啥还用erease这种了,直接一个指针从后往前遍历就AC了。
class Solution { public: int lengthOfLastWord(string s) { int i = s.size()-1; while ( i>=0 ){ if ( s[i]==' ' ) { --i; } else { break; } } int ret = 0; while ( i>=0 ) { if (s[i]!=' ') { ++ret; --i; } else { break; } } return ret; } };