Leetcode 205 同构字符串

　　依然是简单题，熟悉 uthash 的使用。终于想通了，为何封装 HASH_ADD 等相关方法时，需要传入指针的指针，因为 uthash 库的实现都是写在宏中的，编译后不是函数调用，而是代码替换！

　　C：

#include "stdbool.h"
#include <string.h>
#include "./lib/uthash-master/include/uthash.h"

typedef struct Map
{
    char k;
    char v;
    UT_hash_handle hh;
} Map;

bool mapContains(char key, Map **map)
{
    Map *reMap = NULL;
    HASH_FIND(hh, *map, &key, sizeof(char), reMap);
    if (reMap == NULL)
        return false;
    return true;
}

int mapAdd(char key, char value, Map **map)
{
    Map *m = (Map *)malloc(sizeof(Map));
    m->k = key;
    m->v = value;
    if (mapContains(key, map))
        return false;
    HASH_ADD(hh, *map, k, sizeof(char), m);
    return true;
}

char mapGet(char key, Map **map)
{
    Map *tmp = NULL;
    HASH_FIND(hh, *map, &key, sizeof(char), tmp);
    return tmp->v;
}

void MapClear(Map **map)
{
    struct Map *current, *tmp;
    HASH_ITER(hh, *map, current, tmp)
    {
        HASH_DEL(*map, current);
        free(current);
    }
}

bool isIsomorphic(char *s, char *t)
{
    int len = strlen(s);
    Map *dict0 = NULL, *dict1 = NULL;
    bool re = true;
    for (int i = 0; i < len; i++)
    {
        char char0 = s[i], char1 = t[i];
        if ((mapContains(char0, &dict0) && mapGet(char0, &dict0) != char1) || (mapContains(char1, &dict1) && mapGet(char1, &dict1) != char0))
        {
            re = false;
            break;
        }
        mapAdd(char0, char1, &dict0);
        mapAdd(char1, char0, &dict1);
    }
    MapClear(&dict0);
    MapClear(&dict1);
    return re;
}

　　JAVA：

public final boolean isIsomorphic(String s, String t) {
        Map<Character, Character> dict0 = new HashMap<Character, Character>();
        Map<Character, Character> dict1 = new HashMap<Character, Character>();
        for (int i = 0; i < s.length(); i++) {
            char char0 = s.charAt(i), char1 = t.charAt(i);
            if (dict0.keySet().contains(char0) && dict0.get(char0) != char1) return false;
            if (dict1.keySet().contains(char1) && dict1.get(char1) != char0) return false;
            dict0.put(char0, char1);
            dict1.put(char1, char0);
        }
        return true;
    }

　　JS：

/**
 * @param {string} s
 * @param {string} t
 * @return {boolean}
 */
var isIsomorphic = function (s, t) {
    let dict0 = new Map(), dict1 = new Map();
    for (let i = 0; i < s.length; i++) {
        let char0 = s.charAt(i), char1 = t.charAt(i);
        if (dict0.get(char0) && dict0.get(char0) != char1) return false;
        if (dict1.get(char1) && dict1.get(char1) != char0) return false;
        dict0.set(char0, char1);
        dict1.set(char1, char0);
    }
    return true;
};