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LeetCode題目翻譯 英文原文如下
Given a string s consisting of some words separated by some number of spaces, return the length of the last word in the string.
A word is a maximal substring consisting of non-space characters only.
中文翻譯
給予一個由幾個單字且有幾個空白組成的字串 s ,回傳最後一個單字的長度
一個單字的定義是其中只有不含空白之字元
s的長度會介於1到104 之間
s 只會包含英文字元跟空白字元
s 至少會包含一個單字 ➔ 不用考慮s整個為空字串的情況
接著來看看以下的範例
先看範例1
各位應該一眼就能看到最熟悉的兩個單字吧,分別是「Hello」跟「World」這兩個單字吧
故回傳最後的單字「World」的長度 ➔ 5
再來看看範例2
最後一個單字是moon(後面的也是空白故不算)
故回傳最後的單字「moon」的長度 ➔ 4
解題思路C# 解決方案 方案1 – 簡單的遞迴
public class Solution {
public int LengthOfLastWord(string s) {
int cnt = 0;
for(int i = s.Length-1;i>=0;i--)
{
if(Char.IsWhiteSpace(s[i]))
{
if(cnt>0)
{
return cnt;
}
continue;
}
cnt+=1;
}
return cnt;
}
}
最直覺的方式,因為要找最後一個單字的長度,所以從string的最後往前找
方案2 – String 底下的 Split
public class Solution {
public int LengthOfLastWord(string s) {
string[] tmp = s.Split(' '); //使用Split對字串做空白切割
string tmpStr;
for(int i = tmp.Length-1; i>=0; i--) //因為要最後的單字,故從陣列最後一位開始找
{
tmpStr = tmp[i];
if(!String.IsNullOrEmpty(tmpStr)) //不是空的話
{
return tmpStr.Length;
}
}
return 0; //應該不會跑到,題目有保證至少有一個單字
}
}
大家可以看到上面的第三個註解那裡的程式 (空白那個),其實最早寫的時候是沒有的
但實際跑過之後發現,要是最後單字後面有一個或以上的空白,就會導致Array的最後出現空白字元
故還需加上判斷是否為Null或空白(String.IsNullOrEmpty)
方案2.1
已經知道單純的Split有空白的問題,那能不能找到其他辦法解決呢?
當然有! 大家看看上面那一句「最後單字後面有一個或以上的空白」
那我們對症下藥,就先對單字做個預先的處理
這時候我們就會找到內建的String. Trim( ) 這個方法,其中又分為以下3種
Trim( ) TrimEnd( ) TrimStart( ) Trim這個字在英文的意思就是 「修剪」
故方法即為針對字串進行特定字元(若無設定即對象為空白)的修剪
TrimStart 就是對字串的開頭 、TrimEnd 則是對字串的結尾 、Trim則是兩者都做
那現在把它實際運用在剛剛的程式碼中
public class Solution {
public int LengthOfLastWord(string s) {
s = s.TrimEnd(); //Trim也可以,因為這裡只要對結尾做處理即可
string[] tmp = s.Split(' ');
return tmp[tmp.Length-1].Length;
}
}
是不是比剛剛簡單了許多,連迴圈都不用了
而一切就都只是多了一個小小的前置動作而已
方案2.2
什麼! 這樣你們也不滿意!! 不想對字串先處理??? (比案子遇上的客戶還麻煩餒…)
這時候我們要來看下官方對於Split的文件 了
疑?
Split 後面有一個 StringSplitOptions 參數可以帶入的
而其中的「RemoveEmptyEntries」 可以直接省略結果包含空字串的陣列元素
這..這不就我們要的嗎!
事不宜遲,趕緊來改看看
public class Solution {
public int LengthOfLastWord(string s) {
string[] tmp = s.Split(' ',StringSplitOptions.RemoveEmptyEntries);
return tmp[tmp.Length-1].Length;
}
}
Java解 決方案方案1
class Solution {
public int lengthOfLastWord(String s) {
int cnt = 0;
for(int i = s.length()-1;i>=0;i--)
{
if(s.charAt(i)==' ')
{
if(cnt>0)
{
return cnt;
}
continue;
}
cnt+=1;
}
return cnt;
}
}
方案2
class Solution {
public int lengthOfLastWord(String s) {
s = s.trim();
String[] arr = s.split("\\s+");
return arr[arr.length-1].length();
}
}
Python3 解決方案方案1
class Solution:
def lengthOfLastWord(self, s: str) -> int:
cnt = 0;
for i in range(len(s)-1,-1,-1):
if(s[i]== ' '):
if cnt>0 :
return cnt
continue
cnt+=1;
return cnt
⭐方案2
class Solution:
def lengthOfLastWord(self, s: str) -> int:
s = s.rstrip() #or strip
return len(s.split()[-1])
JavaScript 解決方案方案1
/**
* @param {string} s
* @return {number}
*/
var lengthOfLastWord = function(s) {
var cnt = 0;
for(var i = s.length-1;i>=0;i--)
{
if(s[i]==' ')
{
if(cnt>0)
{
return cnt;
}
continue;
}
cnt+=1;
}
return cnt;
};
方案2
/**
* @param {string} s
* @return {number}
*/
var lengthOfLastWord = function(s) {
s = s.trim();
var arr = s.split(' ');
return arr[arr.length-1].length;
};
結論 看完這題各種不同的解法,大家可以看到善用內建的方法會有多少程式碼的差異
而這就是程式之所以讓人著迷的魅力啊
"Always code as if the guy who ends up maintaining your code will be a violent psychopath who knows where you live"
題目連結 : Length of Last Word – LeetCode
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