LeetCode:Length of Last Word

Title:

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（1）：从前往后，设置两个变量。可能出错的地方在于s[i] == ' ',之前没有添加 if (len != 0)这个判断

class Solution {
public:
int lengthOfLastWord(string s) {
if (s.size() == 0)
return 0;
int pre_len = 0;
int len = 0;
for (int i = 0 ; i < s.size(); i++){
if (s[i] == ' '){
if (len != 0)
pre_len = len;
len = 0;
}else{
len++;
}
}
if (len == 0)
return pre_len;
else
return len;
}
};

思路（2）：从后往前。首先去除最后面是空格的，然后往前知道在遇到空格

class Solution {
public:
int lengthOfLastWord(const char *s) {
int len=strlen(s);
int sum=0;
while(s[len-1]==' ') len--;
for(int i=len-1;i>=0;i--)
{
if(s[i]!=' ')   sum++;
else break;
}
return sum;
}
};