【LeetCode】Keyboard Row 解题报告

【LeetCode】Keyboard Row 解题报告

标签（空格分隔）： LeetCode

题目地址：https://leetcode.com/problems/keyboard-row/#/description

题目描述：

Given a List of words, return the words that can be typed using letters of alphabet on only one row’s of American keyboard like the image below.

Example :

Input: [“Hello”, “Alaska”, “Dad”, “Peace”]

Output: [“Alaska”, “Dad”]

Ways

暴力解决了。分别把三行弄在三个数组里，对于每个单词每个字母都去循环，数在三行中的个数分别多少。如果这个单词能在一张中打出来完，那么说明由某一行的个数为1，其他行都为0.

注意字符串数组的写法。

public class Solution {
public String[] findWords(String[] words) {
char []arr1 = new char[]{'q','w','e','r','t','y','u','i','o','p','Q','W','E','R','T','Y','U','I','O','P'};
char []arr2 = new char[]{'a','s','d','f','g','h','j','k','l','A','S','D','F','G','H','J','K','L'};
char []arr3 = new char[]{'z','x','c','v','b','n','m','Z','X','C','V','B','N','M'};
List<String> ans = new ArrayList<String>();
for(String word: words){
int count1 = 0, count2 = 0, count3 = 0;
for(int i =0; i < word.length(); i++){
for(int j =0; j < arr1.length; j++){
if(word.charAt(i) == arr1[j]){
count1++;
}
}
for(int j =0; j < arr2.length; j++){
if(word.charAt(i) == arr2[j]){
count2++;
}
}
for(int j =0; j < arr3.length; j++){
if(word.charAt(i) == arr3[j]){
count3++;
}
}
}
if((count1 != 0 && count2 == 0 && count3 == 0)
||(count1 == 0 && count2 != 0 && count3 == 0)
||(count1 == 0 && count2 == 0 && count3 != 0)){
ans.add(word);
}
}
String []answer = new String[ans.size()];
for(int i =0; i < ans.size(); i ++){
answer[i] = ans.get(i);
}
return answer;
}
}

Date

2017 年 4 月 2 日