Leetcode58. Length of Last Word
2017-06-03 18:11
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Length of Last Word
1、原题
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.
2、题意解析与思路
这题意思很简单,就是让我们找到一个字符串最后一个单词的长度,单词之间是使用" "隔开的。当不存在最后的一个单词时,返回0。思路很简单,就是用一个for循环,从后
往前进行寻找,但遇到空格时,将此时的长度输出。但是我们要注意到一种特殊的情况,就是如"a "这种,标准答案是1。因此在遇到空格时,我们还要在进行一次判断,就是判断当前字符长度是不是为0,如果为0,则继续向前寻找。
3、代码
public int lengthOfLastWord(String s) { StringBuilder str = new StringBuilder(); //从后往前寻找 for (int i = s.length() - 1; i >= 0; i--) { //当遇到空格时 if (s.charAt(i) == ' ') { //如果长度为0,说明还没有遇到最后的一个单词,继续;否则结束返回长度 if (str.length() != 0) { return str.length(); } else { continue; } } else { //非空格时,继续添加字符 str.append(s.charAt(i)); } } return str.length(); }
58. Length of Last Word111111
58. Length of Last Wor
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