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feat: add solutions to leetcode problem: No.0500
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‎solution/0500-0599/0500.Keyboard Row/README.md

+49-2
Original file line numberDiff line numberDiff line change
@@ -40,16 +40,63 @@
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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43-
```python
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用三个集合存储键盘每一行字母。
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遍历单词列表 `words`，判断每个单词 `word` 转 set 后是否被以上三个集合其中之一包含，若是，将单词添加到结果数组。
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```python
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class Solution:
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def findWords(self, words: List[str]) -> List[str]:
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s1 = set('qwertyuiop')
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s2 = set('asdfghjkl')
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s3 = set('zxcvbnm')
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res = []
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for word in words:
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t = set(word.lower())
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if t <= s1 or t <= s2 or t <= s3:
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# 利用 python set 比较
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res.append(word)
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return res
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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用三个字符串存储键盘每一行字母。
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遍历单词列表 `words`，对于每个单词 `word`：
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1. 分别设置三个计数器，存储当前单词在对应键盘字符串的字母个数；
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2. 遍历 `word` 中的每个字母，如果在对应的键盘字符串中，则对应的计数器加 1；
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3. 单词遍历结束后，判断是否存在一个计数器值与 `word` 长度相同。如果有，说明该单词所有字母都在同一个键盘字符串中，将单词添加到结果数组。
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```java
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class Solution {
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public String[] findWords(String[] words) {
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String s1 = "qwertyuiopQWERTYUIOP";
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String s2 = "asdfghjklASDFGHJKL";
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String s3 = "zxcvbnmZXCVBNM";
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List<String> res = new ArrayList<>();
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for (String word : words) {
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int n1 = 0, n2 = 0, n3 = 0;
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int n = word.length();
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for (int i = 0; i < n; ++i) {
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if (s1.contains(String.valueOf(word.charAt(i)))) {
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++n1;
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} else if (s2.contains(String.valueOf(word.charAt(i)))) {
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++n2;
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} else {
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++n3;
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}
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}
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if (n1 == n || n2 == n || n3 == n) {
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res.add(word);
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}
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}
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return res.toArray(new String[0]);
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}
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}
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```
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### **...**

‎solution/0500-0599/0500.Keyboard Row/README_EN.md

+36-2
Original file line numberDiff line numberDiff line change
@@ -38,13 +38,47 @@
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### **Python3**
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```python
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class Solution:
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def findWords(self, words: List[str]) -> List[str]:
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s1 = set('qwertyuiop')
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s2 = set('asdfghjkl')
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s3 = set('zxcvbnm')
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res = []
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for word in words:
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t = set(word.lower())
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if t <= s1 or t <= s2 or t <= s3:
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res.append(word)
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return res
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```
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### **Java**
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```java
47-
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class Solution {
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public String[] findWords(String[] words) {
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String s1 = "qwertyuiopQWERTYUIOP";
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String s2 = "asdfghjklASDFGHJKL";
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String s3 = "zxcvbnmZXCVBNM";
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List<String> res = new ArrayList<>();
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for (String word : words) {
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int n1 = 0, n2 = 0, n3 = 0;
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int n = word.length();
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for (int i = 0; i < n; ++i) {
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if (s1.contains(String.valueOf(word.charAt(i)))) {
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++n1;
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} else if (s2.contains(String.valueOf(word.charAt(i)))) {
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++n2;
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} else {
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++n3;
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}
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}
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if (n1 == n || n2 == n || n3 == n) {
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res.add(word);
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}
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}
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return res.toArray(new String[0]);
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}
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}
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```
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### **...**

‎solution/0500-0599/0500.Keyboard Row/Solution.java

+18-27
Original file line numberDiff line numberDiff line change
@@ -1,34 +1,25 @@
11
class Solution {
2-
32
public String[] findWords(String[] words) {
4-
if (words == null) {
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return null;
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}
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ArrayList<String> list = new ArrayList<>();
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String[] keyboards = {"qwertyuiop", "asdfghjkl", "zxcvbnm"};
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for (int i = 0; i < words.length; i++) {
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String word = words[i].toLowerCase();
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for (int j = 0; j < keyboards.length; j++) {
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// 先用word首字符确定属于哪一行
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if (keyboards[j].indexOf(word.charAt(0)) > -1) {
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// 判断word字符串所有字符是否都属于同一行
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boolean match = match(keyboards[j], word, list);
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if (match) {
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list.add(words[i]);
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}
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break;
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String s1 = "qwertyuiopQWERTYUIOP";
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String s2 = "asdfghjklASDFGHJKL";
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String s3 = "zxcvbnmZXCVBNM";
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List<String> res = new ArrayList<>();
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for (String word : words) {
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int n1 = 0, n2 = 0, n3 = 0;
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int n = word.length();
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for (int i = 0; i < n; ++i) {
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if (s1.contains(String.valueOf(word.charAt(i)))) {
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++n1;
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} else if (s2.contains(String.valueOf(word.charAt(i)))) {
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++n2;
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} else {
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++n3;
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}
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}
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}
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return list.toArray(new String[list.size()]);
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}
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private boolean match(String keyboard, String word, ArrayList<String> list) {
27-
for (int i = 1; i < word.length(); i++) {
28-
if (keyboard.indexOf(word.charAt(i)) < 0) {
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return false;
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if (n1 == n || n2 == n || n3 == n) {
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res.add(word);
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}
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}
32-
return true;
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return res.toArray(new String[0]);
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}
34-
}
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}

‎solution/0500-0599/0500.Keyboard Row/Solution.py

+10-15
Original file line numberDiff line numberDiff line change
@@ -1,16 +1,11 @@
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class Solution:
2-
def findWords(self, words):
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"""
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:type words: List[str]
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:rtype: List[str]
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"""
7-
8-
Rows={"q":0, "w":0, "e":0, "r":0, "t":0, "y":0, "u":0, "i":0, "o":0, "p":0, "a":1, "s":1, "d":1, "f":1, "g":1, "h":1, "j":1, "k":1, "l":1, "z":2, "x":2, "c":2, "v":2, "b":2, "n":2, "m":2}
9-
10-
Final = []
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for each in words :
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if all([Rows[each[0].lower()]==Rows[ch] for ch in each.lower()]) :
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Final.append(each)
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return Final
2+
def findWords(self, words: List[str]) -> List[str]:
3+
s1 = set('qwertyuiop')
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s2 = set('asdfghjkl')
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s3 = set('zxcvbnm')
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res = []
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for word in words:
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t = set(word.lower())
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if t <= s1 or t <= s2 or t <= s3:
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res.append(word)
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return res

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