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.
第一種思路:判斷映射關係,從前往後遍歷判斷映射關係是否和之前的衝突,要注意需要遍歷兩遍,分爲 S-->T和T-->S,我們可以用hash表存儲映射關係,代碼如下:
class Solution {
public:
bool isIsomorphic(string s, string t) {
unordered_map<char,char> map;
if(!s.size()&&!t.size()) return true;
if(s.size()!=t.size()) return false;
for(int i=0;i<s.size();i++){
if(map.find(s[i])==map.end()) map[s[i]]=t[i];
else if(map[s[i]]!=t[i]){
return false;
}
}
map.clear();
for(int i=0;i<t.size();i++){
if(map.find(t[i])==map.end()) map[t[i]]=s[i];
else if(map[t[i]]!=s[i]){
return false;
}
}
return true;
}
};
第二種思路:用類似位圖的思路,用長度256的數組:index代表的字母,值爲該字母(在s,t中)出現的索引和,遍歷一次,逐個判斷索引和是否相同就行了,代碼如下:
<span style="color:#333333;">class Solution {
public:
bool isIsomorphic(string s, string t) {
/*unordered_map<char,char> map;
if(!s.size()&&!t.size()) return true;
if(s.size()!=t.size()) return false;
for(int i=0;i<s.size();i++){
if(map.find(s[i])==map.end()) map[s[i]]=t[i];
else if(map[s[i]]!=t[i]){
return false;
}
}
map.clear();
for(int i=0;i<t.size();i++){
if(map.find(t[i])==map.end()) map[t[i]]=s[i];
else if(map[t[i]]!=s[i]){
return false;
}
}
return true;*/
int map1[256] = {0}, map2[256] = {0};
for (int i = 0; i < s.size();i++) {
if (map1[s[i]] != map2[t[i]]) return false;
</span><span style="color:#ff0000;">map1[s[i]] += i+1;
map2[t[i]] += i+1 </span><span style="color:#333333;">;
}
return true;
}
};</span>
需要加1,主要是因爲map1和map2初始化爲零,要把它和索引爲零區分開。