【Leetcode长征系列】Length of Last Word
2014-08-08 10:56
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原题:
Given a string s consists of upper/lower-case alphabets and empty space characters
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 =
return
思路:读到空格就开始重新计数,直到读完。需要注意的是,如果最后结尾处有一长串的空格需要考虑到特殊情况。
c++中string转char需要用c_str()函数。char数组转string直接赋值就可以了。
isalpha()函数等价于islower()||isupper()函数的作用。
另注意:无论如何,新开辟的值或者新定义的变量一定要赋初始值!不然之后会出现很多不必要的问题。切记切记。
代码:
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.
思路:读到空格就开始重新计数,直到读完。需要注意的是,如果最后结尾处有一长串的空格需要考虑到特殊情况。
c++中string转char需要用c_str()函数。char数组转string直接赋值就可以了。
isalpha()函数等价于islower()||isupper()函数的作用。
另注意:无论如何,新开辟的值或者新定义的变量一定要赋初始值!不然之后会出现很多不必要的问题。切记切记。
代码:
class Solution { public: int lengthOfLastWord(const char *s) { string s1 = s; if (s1.length()==0) return 0; int n=0, pre=0; for(int i = 0; i<s1.length(); i++){ if(n!=0) pre = n; if(s1[i]==' ') n = 0; else n++; } if(s[s1.length()-1]==' ') return pre; return n; } };
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