一天一道LeetCode#58. Length of Last Word
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一天一道LeetCode系列
(一)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.
(二)解题
题目比较简单,从尾向头找,第一个不为空格的字母,然后再继续找遇到空格为止。纪录单词长度。
class Solution {
public:
int lengthOfLastWord(string s) {
if(s.length() == 0) return 0;
int count = 0;
int i = s.length()-1;
while(s[i]==‘ ‘) i--;//忽略前面的空格
while(i>=0&&s[i]!=‘ ‘) //当为字母的时候计算长度,为空格就退出循环
{
count++;
i--;
}
return count;
}
};
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