算法（三十三）

1、判定字符串B是否可由字符串A变换的来  比如  “abccd”    “acbcd” 可以。

public boolean fun(String a, String b){
    int len1 = a.length();
    int len2 = b.length();
    HashMap<String> hm1 = new HashMap<String>;
    HashMap<String> hm2 = new HashMap<String>;
    boolean flag = true;
    
    if(len1 != len2) (return false;}
    for(int i=0; i<len1; i++){
        if(!hm1.containsKey(a.charAt(i))){
            hm1.put(a.charAt(i), 1);
        }else{
            val = hm1.get(a.charAt(i));
            hm1.put(a.charAt(i), ++val);
        }    
    }
    for(int i=0; i<len2; i++){
        if(!hm2.containsKey(b.charAt(i))){
            hm2.put(b.charAt(i), 1);
        }else{
            val = hm2.get(b.charAt(i));
            hm2.put(b.charAt(i), ++val);
        }    
    }
    
    for(int i=0; i<len1; i++){
        if(hm1.containsKey(b.charAt(i)) && (hm1.get(b.charAt(i)) == hm2.get(b.charAt(i)))){
            continue;
        }else{
            flag = false;
            break;
        }    
    }
    return flag;    
}

2.给定形如pattern="aabbc"，city="北京 北京 上海 上海 蓬莱"两个字符串，这样的pattern和city之间是匹配的（a=北京，b=上海，c=蓬莱）。规定pattern内的字符属于[a-z]，city的城市名必定以空格分隔，给定任意pattern和city，返回它们是否匹配（True/False）

public class test002 {
    private static boolean fun1(String pattern, String city) {
        HashMap<Character, String> hm = new HashMap<>();
        String[] new_city = city.split(" ");
        int len1 = pattern.length();
        int len2 = new_city.length;
        boolean flag = false;
        if(len1 != len2) return false;

        for(int i=0; i<len1; i++){
            hm.put(pattern.charAt(i), new_city[i]);
        }

        for(int i=0; i<len1; i++){
            if(hm.get(pattern.charAt(i)).equals(new_city[i])){
                flag=true;
            }else{
                flag=false;
                break;
            }

        }
        return flag;
    }

    private static boolean fun2(String city, String pattern) {
        HashMap<String, Character> hm = new HashMap<>();
        String[] new_city = city.split(" ");
        int len1 = pattern.length();
        int len2 = new_city.length;
        boolean flag = false;
        if(len1 != len2) return false;

        for(int i=0; i<len1; i++){
            hm.put(new_city[i], pattern.charAt(i));
        }

        for(int i=0; i<len1; i++){
            if(hm.get(new_city[i]).equals(pattern.charAt(i))){
                flag=true;
            }else{
                flag=false;
                break;
            }

        }
        return flag;
    }

    public static void main(String[] args) {
        String pattern= "aabbc";
        String city = "北京 北京 上海 上海 蓬莱";
        boolean flag1 = fun1(pattern, city);
        boolean flag2 = fun2(city, pattern);
        if(flag1 && flag2) System.out.println(true);
        else{
            System.out.println(false);
        }
    }
}