leetcode：Isomorphic Strings

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;
}