205. Isomorphic Strings

题目：

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.

链接： http://leetcode.com/problems/isomorphic-strings/

题解：

一开始的想法就是使用HashMap，试了一下发现必须一一对应，所以想到了类似电话簿，建立两个hashmap来保存这种一一对应的关系。

Time Complexity - O(n)， Space Complexity - O(1)。

public class Solution {
    public boolean isIsomorphic(String s, String t) {
        if(s == null || t == null ||  s.length() != t.length())
            return false;
    
        Map<Character, Character> st = new HashMap<>();
        Map<Character, Character> ts = new HashMap<>();
        
        for(int i = 0; i < s.length(); i++) {
            if(!st.containsKey(s.charAt(i))) {
                st.put(s.charAt(i), t.charAt(i));
            } else {
                if(st.get(s.charAt(i)) != t.charAt(i))
                    return false;
            }
            if(!ts.containsKey(t.charAt(i))) {
                ts.put(t.charAt(i), s.charAt(i));
            } else {
                if(ts.get(t.charAt(i)) != s.charAt(i))
                    return false;
            }
        }
        
        return true;
    }
}

还有更好的写法，就是开一个到两个array，然后比较ASCII，这样应该能算是O(1)和O(1)。二刷要多学习。

二刷:

使用GrandYong的写法，使用两个bitmap，然后存储sChar和tChar的位置，注意 sMap[sChar] = i + 1和tMap[tChar] = i +１是为了区别初始的0。假如我们初始化数组都等于-1的话，就可以不用+1。不过这个也无关紧要

Java:

Time Complexity - O(n)， Space Complexity - O(1)。

public class Solution {
    public boolean isIsomorphic(String s, String t) {
        if (s == null || t == null || s.length() != t.length()) {
            return false;
        }
        int[] sMap = new int[256], tMap = new int[256];
        for (int i = 0; i < s.length(); i++) {
            char sChar = s.charAt(i);
            char tChar = t.charAt(i);
            if (sMap[sChar] != tMap[tChar]) {
                return false;
            }
            sMap[sChar] = i + 1;   // to differential case 0, 0
            tMap[tChar] = i + 1;
        }
        return true;
    }
}

三刷:

Java:

对Unicode还是要用两个hashMap。 Space Complexity，即使是Unicode也只是小常数，所以还是算O(1)。

Time Complexity - O(n)， Space Complexity - O(1)。

public class Solution {
    public boolean isIsomorphic(String s, String t) {
        if (s == null || t == null || s.length() != t.length()) {
            return false;
        }
        Map<Character, Character> st = new HashMap<>();
        Map<Character, Character> ts = new HashMap<>();
        
        for (int i = 0; i < s.length(); i++) {
            char sChar = s.charAt(i);
            char tChar = t.charAt(i);
            if ((st.containsKey(sChar) && st.get(sChar) != tChar) || (ts.containsKey(tChar) && ts.get(tChar) != sChar)) {
                return false;
            } 
            if (!st.containsKey(sChar)) {
                st.put(sChar, tChar);    
            }
            if (!ts.containsKey(tChar)) {
                ts.put(tChar, sChar);    
            }
        }
        return true;
    }
}

假设Alphabet为Single-byte char，我们就可以使用bitmap，速度大大提高。这里是对每个元素的出现位置进行比较，并且使用的是 1- based index来避免 array数组初始化的0。每次遇到两个字符，我们就去sMap和tMap中来比较它们上一次的出现的位置是否相同，假如不通的话我们就返回false。

Time Complexity - O(n)， Space Complexity - O(1)。

public class Solution {
    public boolean isIsomorphic(String s, String t) {
        if (s == null || t == null || s.length() != t.length()) {
            return false;
        }
        int[] sMap = new int[256];
        int[] tMap = new int[256];
        
        for (int i = 0; i < s.length(); i++) {
            char sChar = s.charAt(i);
            char tChar = t.charAt(i);
            if (sMap[sChar] != tMap[tChar]) {
                return false;
            }
            sMap[sChar] = i + 1;
            tMap[tChar] = i + 1;
        }
        return true;
    }
}

Reference:

https://leetcode.com/discuss/33889/short-java-solution-without-maps

https://leetcode.com/discuss/55777/fast-java-solution-using-ascii-code

205. Isomorphic Strings

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