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题目描述：
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.
解题思路：
先定位到字符串从后往前数第一个不为空字符的位置，然后进行判断。
从第一个位置向后读取，每遇到空字符则将计数器清零，然后下次继续计数。
代码如下：
public class Solution {
public int lengthOfLastWord(String s) {
int curLen = 0, len = 0, end = -1;
if(s == null || (len = s.length()) == 0)
return 0;
for(int i = len - 1; i >= 0; i--){
if(s.charAt(i) != ‘ ‘) {
end = i;
break;
}
}
for(int i = 0; i <= end; i++){
if(s.charAt(i) != ‘ ‘)
curLen++;
else {
curLen = 0;
}
}
return curLen;
}
}
Java [Leetcode 58]Length of Last Word
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原文地址：http://www.cnblogs.com/zihaowang/p/5183653.html
