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LeetCode Day 34: P205. Isomorphic Strings (Easy)

题目：

两个字符串，s和t，判断它们是否同态，一一映射。

Input: s = "egg", t = "add"
Output: true

Input: s = "foo", t = "bar"
Output: false

我的思路：

我原以为只有小写字母，所以就想，不用Hashmap, 通过ASCII码天然自带的映射关系，判定它们是否出现过，如果出现过，断定 它们是否是正确的对应字母。

遇到的一个问题就是需要判定两遍，分别用s和t作为数组的索引，另一个作为判定的字符。

class Solution {
    public boolean isIsomorphic(String s, String t) {
        if (s.length() != t.length()) {
            return false;
        }
        char[] arr_s = new char[127];
        for (int i = 0; i < s.length(); i++) {
            char current_t = arr_s[s.charAt(i)];
            if ( current_t == 0) {
                arr_s[s.charAt(i)] = t.charAt(i);
            } else if (current_t != t.charAt(i)){
                return false;
            }
        }

        char[] arr_t = new char[127];
        for (int i = 0; i < t.length(); i++) {
            char current_s = arr_t[t.charAt(i)];
            if ( current_s == 0) {
                arr_t[t.charAt(i)] = s.charAt(i);
            } else if (current_s != s.charAt(i)){
                return false;
            }
        }
        return true;
    }
}

结果还算可以，87.5%

最优解

class Solution {
    public boolean isIsomorphic(String s, String t) {

        if (s == null || s.isEmpty())
            return (t == null || t.isEmpty());

        if (s.length() != t.length())
            return false;

        char[] s1 = s.toCharArray();
        char[] t1 = t.toCharArray();

        int length = s.length();

        char[] sm = new char[256];
        char[] tm = new char[256];

        for (int i = 0; i < length; i++) {
            char sc = s1[i];
            char tc = t1[i];

            /**
             * Have u seen sc and tc before?
             */
            if (sm[sc] == 0 && tm[tc] == 0) {
                /**
                 * If not, create a mapping from sc -> tc
                 */
                sm[sc] = tc;
                tm[tc] = sc;
            } else {
                /**
                 * If seen, then they must be same mapping
                 */
                if (sm[sc] != tc || tm[tc] != sc) {
                    return false;
                }
            }
        }
        return true;
    }
}

看完代码，感觉完全像我的升级版，思路比我清楚多了，而且一趟跑完。

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