Leetcode 205 Isomorphic Strings 同構字符串

原題地址

https://leetcode.com/problems/isomorphic-strings/

題目描述

Given two strings s and t, determine if they are isomorphic.
給出兩個字符串s和t，判斷它們是否是同構的。

Two strings are isomorphic if the characters in s can be replaced to get t.
如果可以通過替換s中的字符來得到t，則兩個字符串是同構的。

All occurrences of a character must be replaced with another character while preserving the order of characters. No two characters may map to the same character but a character may map to itself.
同一個字符只能替換成另外一個字符，且兩個字符不能映射成同一個字符，但是字符可以映射爲自身（也就是沒有替換）。

For example,
例如，

Given，
給出

s = "egg", t = "add", return true.
s = "foo", t = "bar", return false.
s = "paper", t = "title", return true.

Tags Hash Table

解題思路

這個題比較簡單，嘗試將s[i]替換爲t[i]，記錄一些從s中的字符替換到t之後的映射以及t中字符對s中字符的映射，然後在嘗試替換時看與之前的映射是否有衝突即可。詳見代碼。

另外需要注意的是，一個字符爲8位，大小界於0~255之間。

代碼

/** 判斷兩個字符串是否同構 */
bool isIsomorphic(char* s, char* t) {   
    int len = strlen(s);
    if (len != strlen(t)) return false;

    // mapTo表示s中的某種字符替換到t之後的字符
    // mapFrom表示t中的某種字符替換到s之後字符
    int mapTo[256], mapFrom[256];
    // 初始化爲-1
    int i = -1;
    while (++i < 256) mapTo[i] = mapFrom[i] = -1;

    i = 0;
    char from, to;
    while (i < len) {
        from = *(s + i);
        to = *(t + i);
        // 如果兩個字符都沒有做過替換
        if (mapTo[from] == -1 && mapFrom[to] == -1) { 
            mapTo[from] = to;  // 替換並記錄兩個字符的映射
            mapFrom[to] = from;
        } else {
            // 如果兩個字符中有一個做過映射，且與當前映射情況不符，返回false
            if (mapTo[from] != -1 && mapTo[from] != to)
                return false;
            if (mapFrom[to] != -1 && mapFrom[to] != from)
                return false;
        }
        ++i;
    }

    return true;
}

完整代碼　https://github.com/Orange1991/leetcode/blob/master/205/c/main.c

測試數據

s = egg,
t = add,
they are isomorphic.

s = foo,
t = bar,
they are not isomorphic.

s = paper,
t = title,
they are isomorphic.

s = a,
t = a,
they are isomorphic.

s = ab,
t = ab,
they are isomorphic.

s = 13,
t = 42,
they are isomorphic.

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2015/8/7