【Leetcode】500. Keyboard Row

題目地址：

https://leetcode.com/problems/keyboard-row/

給定一個字符串數組，返回其中可以由鍵盤中其中一行打出來的所有字符串。

先把鍵盤的每一行存入三個哈希表，然後遍歷數組。可以用一個數$x$的二進制位來記錄每個字符串用了那幾行，然後判斷一下$x$是否等於$1$、$2$或者$8$即可。代碼如下：

import java.util.*;

public class Solution {
    public String[] findWords(String[] words) {
        List<String> ans = new ArrayList<>();
        Set<Character>[] sets = (HashSet<Character>[]) new HashSet[3];
        String[] rows = {"qwertyuiop", "asdfghjkl", "zxcvbnm"};
        for (int i = 0; i < rows.length; i++) {
            sets[i] = new HashSet<>();
            for (int j = 0; j < rows[i].length(); j++) {
                sets[i].add(rows[i].charAt(j));
            }
        }
    
        for (int i = 0; i < words.length; i++) {
            int row = 0;
            for (int j = 0; j < words[i].length(); j++) {
                char ch = Character.toLowerCase(words[i].charAt(j));
                // 如果發現ch在其中某一行，就將那一行的編號的二進制位置爲1
                if (sets[0].contains(ch)) {
                    row |= 1;
                }
                if (sets[1].contains(ch)) {
                    row |= 2;
                }
                if (sets[2].contains(ch)) {
                    row |= 8;
                }
            }
            
            if (row == 1 || row == 2 || row == 8) {
                ans.add(words[i]);
            }
        }
        
        String[] res = new String[ans.size()];
        for (int i = 0; i < res.length; i++) {
            res[i] = ans.get(i);
        }
        
        return res;
    }
}

時間複雜度$O(nl)$，空間$O(nl)$。