一天一道LeetCode#58. Length of Last Word

Posted ZeeCoder

tags:

篇首语:本文由小常识网(cha138.com)小编为大家整理,主要介绍了一天一道LeetCode#58. Length of Last Word相关的知识,希望对你有一定的参考价值。

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;

    }

};

以上是关于一天一道LeetCode#58. Length of Last Word的主要内容,如果未能解决你的问题,请参考以下文章

一天一道LeetCode#62. Unique Paths

一天一道LeetCode#57. Insert Interval

一天一道LeetCode#49. Group Anagrams

一天一道LeetCode#52. N-Queens II

一天一道LeetCode#60. Permutation Sequence.

一天一道LeetCode#63. Unique Paths II