leetcode -- 58. Length of Last Word【遍历数组的次序：前后】

题目

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

.

题意

给定一个字符串 ，返回这个字符串最后一个单词的长度。

分析及解答

解答1：（从后往前）【更快】

public class Solution {
public int lengthOfLastWord(String s) {
if(s.equals("")) return 0;
char[] array = s.toCharArray();

boolean isPreSpace = true;
boolean isCurrentSpace = false;
int count = 0;

for(int i = array.length-1; i >= 0 ;i--){
isCurrentSpace = (array[i] == ' ');

if(isPreSpace){
if(!isCurrentSpace){
count =1;
}
}else{
if(!isCurrentSpace){
count ++;
}else{
break;
}
}
isPreSpace = isCurrentSpace;
}
return count;
}
}

解法2：（从前往后）

public int lengthOfLastWord(String s) {
if(s.equals("")) return 0;
char[] array = s.toCharArray();

boolean isPreSpace = true;
boolean isCurrentSpace = false;
int count = 0;

for(char ch : array){
isCurrentSpace = (ch == ' ');

if(isPreSpace){
if(!isCurrentSpace){
count =1;
}
}else{
if(!isCurrentSpace){
count ++;
}
}
isPreSpace = isCurrentSpace;
}
return count;
}