Go语言LeetCode500键盘行题解示例详解

输入：words = ["Hello","Alaska","Dad","Peace"]
输出：["Alaska","Dad"]

输入：words = ["omk"]
输出：[]

输入：words = ["adsdf","sfd"]
输出：["adsdf","sfd"]

class Solution {
    public String[] findWords(String[] words) {
        String[] res = new String[words.length];
        String fir = "qwertyuiopQWERTYUIOP";
		String sec = "asdfghjklASDFGHJKL";
		String tir = "zxcvbnmZXCVBNM";
        int k = 0;
        for(int i = 0;i<words.length;i++){
            String temp = "";
            for(int j = 0;j<words[i].length();j++){
                if(fir.indexOf(words[i].charAt(0))!=-1){
                    temp = fir;
                }else if(sec.indexOf(words[i].charAt(0))!=-1){
                    temp = sec;
                }else{
                    temp = tir;
                }
                if(temp.indexOf(words[i].charAt(j))==-1){
                    break;
                }
                if(temp.indexOf(words[i].charAt(j))!=-1&&j==words[i].length()-1){
                    res[k]=words[i];
                    k++;
                }
            }
        }
        String[] result = new String[k];
        for(int p = 0;p<k;p++){
            result[p] = res[p];
        }
        return result;
    }
}

            bool b1 = true;
            bool b2 = true;
            bool b3 = true;
            for (auto c: word) {
                b1 &= m1[c];
                b2 &= m2[c];
                b3 &= m3[c];
            }            
            if (b1 || b2 || b3) ans.push_back(word);

class Solution {
public:
    string one = "qwertyuiopQWERTYUIOP";
    string two = "asdfghjklASDFGHJKL";
    string three = "zxcvbnmZXCVBNM";
    unordered_map<char, int> m1,m2,m3;
    vector<string> findWords(vector<string>& words) {
        for (auto c: one) {
            m1[c]++;
        }
        for (auto c: two) {
            m2[c]++;
        }
        for (auto c: three) {
            m3[c]++;
        }
        vector<string> ans;
        for (auto word: words) {
            bool b1 = true;
            bool b2 = true;
            bool b3 = true;
            for (auto c: word) {
                b1 &= m1[c];
                b2 &= m2[c];
                b3 &= m3[c];
            }
            
            if (b1 || b2 || b3) ans.push_back(word);
        }
        return ans;
    }
};