389. Find the Difference

Given two strings s and t which consist of only lowercase letters.

String t is generated by random shuffling string s and then add one more letter at a random position.

Find the letter that was added in t.

Example:

Input:
s = "abcd"
t = "abcde"

Output:
e

Explanation:
‘e‘ is the letter that was added.

我的做法, hashMap, O(n) space, O(n) time:

public class Solution {
    public char findTheDifference(String s, String t) {
        if (t.length() == 1) {
            return t.charAt(0);
        }
        Map<Character, Integer> set = new HashMap<>();
        Map<Character, Integer> map = new HashMap<>();
        for (int i = 0; i < s.length(); i++) {
            set.put(s.charAt(i), set.getOrDefault(s.charAt(i), 0) + 1);
            map.put(t.charAt(i), map.getOrDefault(t.charAt(i), 0) + 1);
        }
       map.put(t.charAt(t.length() - 1), map.getOrDefault(t.charAt(t.length() - 1), 0) + 1);
        for (int i = 0; i < t.length(); i++) {
            if (!set.containsKey(t.charAt(i)) || set.get(t.charAt(i)) < map.get(t.charAt(i))) {
                return t.charAt(i);
            }
        }
        return ‘1‘;
    }
}

用ascii 码表, 时间, 空间都是O(1) 学会转化: (int) s.charAt(i)

public class Solution {
    public char findTheDifference(String s, String t) {
        // Initialize variables to store sum of ASCII codes for
        // each string
        int charCodeS = 0, charCodeT = 0;
        // Iterate through both strings and char codes
        for (int i = 0; i < s.length(); ++i) charCodeS += (int)s.charAt(i);
        for (int i = 0; i < t.length(); ++i) charCodeT += (int)t.charAt(i);
        // Return the difference between 2 strings as char
        return (char)(charCodeT - charCodeS);
    }
}