标签：
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.
这道题就是给定一个字符串，然后判断最后一个单词的长度。似乎是可以直接调用split()的。。。当然，并不是最优的。但却是最简单的，不过了解一下split()也好
public class Solution { public int lengthOfLastWord(String s) { String[] tt; tt = s.split(" "); return tt.length == 0?0:tt[tt.length-1].length(); } }
换用正经的做法。也很简单，一次循环搞定。然而是我太天真了，从前向后必然不是最优。。。。
public class Solution { public int lengthOfLastWord(String s) { int result = 0,len = s.length(); for( int i = 0;i<len;i++){ if( s.charAt(i) == ‘ ‘){ while( i<s.length() && s.charAt(i) == ‘ ‘) i++; if( i!= len) result = 1; } else result++; } return result; } }
好吧，从后往前，这样就ok了。
public class Solution { public int lengthOfLastWord(String s) { int result = 0,i = s.length()-1; while( i>=0 && s.charAt(i) == ‘ ‘ ){ i--; } while( i>=0 && s.charAt(i) != ‘ ‘){ result++; i--; } return result; } }
leetcode 58 Length of Last Word ----- java
标签：
原文地址：http://www.cnblogs.com/xiaoba1203/p/5808782.html
