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
.
题目解析:
让求最后的单词的长度,本来以为只要碰到空格就更新index,最后用字符串长度len与index之间的差值表示就行。但是忘了一种情况“a ”,在单词最后还有空格,那么就需要特殊处理。本来还想用前后指针i,j不过也没得到很好的思路。后来通过计数count来判断。当碰到空格,并且空格的下一个是字符的时候,才将count清零表示重新计算。并且遇到空格还不能让count++。注意这些情况,就能得到正确的结果了。
class Solution {
public:
int lengthOfLastWord(const char *s) {
int len = strlen(s);
if(s == NULL || len == 0)
return 0;
int count = 0;
if(s[0] != ' ')
count++;
for(int i = 1;i < len;i++)
if(s[i] == ' '){
if(i+1 < len && s[i+1] != ' ')
count = 0;
}else
count++;
return count;
}
};