Isomorphic Strings

Given two strings s and t, determine if they are isomorphic.

Two strings are isomorphic if the characters in s can be replaced to get t.

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.

For example,
Given "egg", "add", return true.

Given "foo", "bar", return false.

Given "paper", "title", return true.

Note:
You may assume both s and t have the same length.

分析：同构字符串必须满足，s字符串中相同的字符对应t字符串中相同的字符，s字符串中不同的字符对应t字符串中不同的字符。

代码如下：

class Solution {

public:

     bool isIsomorphic(string s, string t) {

        char map_s[128] = { 0 };

        char map_t[128] = { 0 };

        int len = s.size();

        for (int i = 0; i < len; ++i)

        {

            if (map_s[s[i]]!=map_t[t[i]]) return false;

            map_s[s[i]] = i+1;

            map_t[t[i]] = i+1;

        }

        return true;

    }

};

可参考解法：

记录遍历s的每一个字母，并且记录s[i]到t[i]的映射，当发现与已有的映射不同时，说明无法同构，直接return false。然后交换s和t的位置再重来一遍，这样保证s和t之间的双向映射。

 class Solution {

  public:

     bool isIsomorphic(string s, string t) {

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

          map<char, char> mp;

          for (int i = 0; i < s.length(); ++i) {

             if (mp.find(s[i]) == mp.end()) mp[s[i]] = t[i];

             else if (mp[s[i]] != t[i]) return false;

          }

         mp.clear();

         for (int i = 0; i < s.length(); ++i) {

             if (mp.find(t[i]) == mp.end()) mp[t[i]] = s[i];

             else if (mp[t[i]] != s[i]) return false;

         }

         return true;

      }

 };

题意为判断一个字符串中是否可以由另一个字符串中的字符替换而来。

如果直接尝试替换的话实现比较麻烦，可以对两个字符串分解进行转换，看转换后的结果是否一致。　

依次用‘0’， ‘1‘...替换字符串出现的字符，如 ’abbc‘可以替换为’0112‘。需要设置一张转换表，记录转换后每个字符对应的替代字符：

class Solution {

public:

     string transferStr(string s){

        char table[128] = {0};

        char tmp = '0';

        for (int i=0; i<s.length(); i++) {

            char c = s.at(i);

            if (table[c] == 0) {

                table[c] = tmp++;

            }

            s[i] = table[c];

        }

        return s;

    }

    bool isIsomorphic(string s, string t) {  

        if (s.length() != t.length()) {

            return false;

        }

        if (transferStr(s) == transferStr(t)) {

            return true;

        }

        return false;

    }

};

其他：

class Solution {

public:

    bool isIsomorphic(string s, string t)

    {

        const size_t n = s.size();

        if ( n != t.size())

            return false;

        unsigned char forward_map[256] = {}, reverse_map[256] = {};

        for ( int i=0; i < n; ++i)

        {

            unsigned char c1 = s[i];

            unsigned char c2 = t[i];

            if ( forward_map[c1] && forward_map[c1] != c2)

                return false;

            if ( reverse_map[c2] && reverse_map[c2] != c1)

                return false;

            forward_map[c1] = c2;

            reverse_map[c2] = c1;

        }

        return true;

    }

};

3 lines 3ms C solution

bool isIsomorphic(char* s, char* t) {

static char n[512],*m = n + 256;

return (!*s && !*t && memset(n,0,512)) || (((!(m[*s] || n[*t] || !(m[*s] = *t, n[*t] = *s)) ||

(m[*s] == *t && n[*t] == *s)) || !memset(n,0,512))  && isIsomorphic(s+1,t+1)); }

3ms O(n) C solution：The idea is to use two arrays to store the mappings between corresponding characters of s and t.

bool isIsomorphic(char* s, char* t) {

    char map[256], rmap[256];

    memset(map, 0, sizeof map);

    memset(rmap, 0, sizeof rmap);

    for ( ; *s; ++s, ++t)

        if (!map[*s]) {

            if (rmap[*t]) // another character already maps to *t

                return false;

            map[*s] = *t;

            rmap[*t] = *s;

        } else if (map[*s] != *t)

            return false;

    return true;

}

巴特西

leetcode：Isomorphic Strings

Isomorphic Strings

其他：

3ms O(n) C solution：The idea is to use two arrays to store the mappings between corresponding characters of s and t.

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