LeeCode 58. Length of Last Word 计算字符串的最后的一个单词的长度
2016-10-31 15:39
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题目要求:
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.
说明:
记住使用 s.trim()方法用于将字符串前后的空格去掉
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.
说明:
记住使用 s.trim()方法用于将字符串前后的空格去掉
public class solution { public int lengthOfLastWord(String s) { //使用数组的方式进行处理 s = s.trim(); // s.trim() 除去字符串前后的所有的空格 if(s.isEmpty()) return 0; String[] strarr = s.split(" "); String strx= strarr[strarr.length - 1]; return strarr[strarr.length - 1].length(); } }
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