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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：
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;
}
};
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)); }
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;
}
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原文地址：http://www.cnblogs.com/carsonzhu/p/4639652.html
