LeetCode: Length of Last Word
2014-06-25 08:56
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思路：这个题的输入比较简单，只包含字母和空字符，省去了其他字符的考虑，唯一需要考虑的是末尾如果出现一个或多个空字符就返回注意前面出现的单词大小。
code:
code:
class Solution {
public:
bool isLetter(char a){
return ((a >= 'a' && a <= 'z') || (a >= 'A' && a <= 'Z')) ? 1 : 0;
}
int lengthOfLastWord(const char *s) {
int len = strlen(s), leftNum = 0, rightNum = 0;
for(int i = 0;i<len;i++){
if(isLetter(s[i])){
leftNum++;
rightNum = leftNum;
}
else if(s[i] == ' ')
leftNum = 0;
}
return rightNum;
}
};
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