Length of Last Word leetocde java
2017-08-24 22:52
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题目:
Given a string s consists of upper/lower-case alphabets and empty space characters
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 =
return
题解:
这道题主要是考虑一下最后是不是空格,方法是倒着找不是空格的字符并计数,如果遇到空格且计数不是0,说明最后一个单词已经被计数了,所以可以返回了。
代码如下:
public int lengthOfLastWord(String s) {
if (s == null || s.length() == 0)
return 0;
int len = s.length();
int count = 0;
for (int i = len - 1; i >= 0; i--) {
if (s.charAt(i) != ' ') {
count++;
}
if(s.charAt(i)==' '&&count != 0){
return count;
}
}
return count;
}
当然这道题也能用投机取巧的方法,用split函数把字符串按照空格分隔好,返回最后那个就行。。。
代码如下:
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.
题解:
这道题主要是考虑一下最后是不是空格,方法是倒着找不是空格的字符并计数,如果遇到空格且计数不是0,说明最后一个单词已经被计数了,所以可以返回了。
代码如下:
public int lengthOfLastWord(String s) {
if (s == null || s.length() == 0)
return 0;
int len = s.length();
int count = 0;
for (int i = len - 1; i >= 0; i--) {
if (s.charAt(i) != ' ') {
count++;
}
if(s.charAt(i)==' '&&count != 0){
return count;
}
}
return count;
}
当然这道题也能用投机取巧的方法,用split函数把字符串按照空格分隔好,返回最后那个就行。。。
代码如下:
public int lengthOfLastWord(String s) { String[] a = s.split(" "); if(a == null || a.length == 0) return 0; return a[a.length-1].length(); }
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