Leetcode:205. Isomorphic Strings(JAVA)
2016-03-15 17:24
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【问题描述】
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
return true.
Given
return false.
Given
return true.
【思路】
保存一个hashmap,保存映射关系
同时维护一个set,保证一个character不能被同时映射到两个字符上。
【code】
public class Solution {
public boolean isIsomorphic(String s, String t) {
if (s.length() != t.length()) {
return false;
}
HashMap<Character, Character> map = new HashMap<Character, Character>();
Set<Character> set = new HashSet<Character>();
for (int i = 0; i < s.length(); i++) {
char sc = s.charAt(i);
char tc = t.charAt(i);
if (map.containsKey(sc)) {
if (map.get(sc) == tc) {
continue;
} else {
return false;
}
}else {
if (set.contains(tc)) {
return false;
}else {
map.put(sc, tc);
set.add(tc);
}
}
}
return true;
}
}
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.
【思路】
保存一个hashmap,保存映射关系
同时维护一个set,保证一个character不能被同时映射到两个字符上。
【code】
public class Solution {
public boolean isIsomorphic(String s, String t) {
if (s.length() != t.length()) {
return false;
}
HashMap<Character, Character> map = new HashMap<Character, Character>();
Set<Character> set = new HashSet<Character>();
for (int i = 0; i < s.length(); i++) {
char sc = s.charAt(i);
char tc = t.charAt(i);
if (map.containsKey(sc)) {
if (map.get(sc) == tc) {
continue;
} else {
return false;
}
}else {
if (set.contains(tc)) {
return false;
}else {
map.put(sc, tc);
set.add(tc);
}
}
}
return true;
}
}
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