Easy
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.
5ms:
public int lengthOfLastWord(String s) {
int i=0,j=0;
while(j<s.length()){
if(s.charAt(s.length()-1-j)==' ')
j++;
else
break;
}
while(i+j<s.length()){
if(s.charAt(s.length()-1-i-j)==' ')
return i;
i++;
}
return i;
}
精简:
public int lengthOfLastWord(String s) {
int len = s.length() - 1;
while(len > 0 && s.charAt(len) == ' ') len--;
int ans = 0;
for (int i = len; i >= 0; --i ){
if (s.charAt(i) == ' ') break;
ans++;
}
return ans;
}
本文介绍了一种高效的方法来计算给定字符串中最后一个非空格字符序列(单词)的长度。通过从字符串末尾开始遍历并忽略尾部空格,接着继续向前直到遇到空格为止,从而确定最后一个单词的长度。
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