389. Find the Difference

最新推荐文章于 2022-07-23 09:38:49 发布

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本文介绍了一种通过比较两个字符串来找出其中一个字符串相对于另一个字符串新增加的字符的方法。使用了两种不同的解决方案：一种是通过统计字符出现次数的差异来找到新增的字符；另一种则是利用位操作符进行异或运算，巧妙地简化了问题，直接找到了新增的字符。

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.

My solution:

public class Solution {
    public char findTheDifference(String s, String t) {
        int[] sum_s=new int[26];
        int[] sum_t=new int[26];
        int num_s=s.length();
        int num_t=t.length();
        char c;
        int i;
        for(i=0;i<num_s;++i){
            int index=s.charAt(i)-97;
            sum_s[index]++;
            }
        for(i=0;i<num_t;++i){
            int index=t.charAt(i)-97;
            sum_t[index]++;
            }
        for(i=0;i<26;++i){
            if(sum_s[i]!=sum_t[i]){
                i=i+97;
                break;
            }
            }
        c=(char) i;    
        return c;    
        
    }
}

a better solution with bit manipulation:

a xor a=0;

a xor 0=a;

a xor b=b xor a;

(a xor b) xor c=a xor (b xor c);

public class Solution {
    public char findTheDifference(String s, String t) {
        int len_s=s.length();
        int len_t=t.length();
        char c=t.charAt(len_t-1);
        for(int i=0;i<len_t-1;++i){
            c^=s.charAt(i);
            c^=t.charAt(i);
            }
        return c;    
        
    }
}