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  • 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);
        }
    }
    

      

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  • 原文地址:https://www.cnblogs.com/apanda009/p/7143267.html
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