[LeetCode]389. Find the Difference

依旧是利用a^b^b==a的bit manipulation的特性。另一个方法则利用了将字母作为数组下标的方法。（从本质上来讲，在计算机里面，可以认为char就是int了，就是说可以把char当作int来用，如，作为数组下标。）

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.

方法一：用HashMap

public class Solution {
public char findTheDifference(String s, String t) {
//char res='';
Map<Character,Integer> map = new HashMap<>();
for(char c : s.toCharArray())
map.put(c,map.getOrDefault(c,0)+1);

for(char c : t.toCharArray())
{
map.put(c,map.getOrDefault(c,0)-1);
if(map.get(c)==-1)
return c;
}

return 0;
}
}

方法二：用数组。本质上和方法一相同。

for (int i = 0; i < 26; i++) alpha[i] = 0;
for (char c : s.t
96eb
oCharArray())
alpha[ c - 'a' ]++;

for (char c : t.toCharArray()) {
//could do decrement first, then check but yeah
if (--alpha[c - 'a'] < 0)
return c;
}

return 0;

一定要学会 arr[ c - 'a' ]++ 这种用法。

方法三：异或。a^b^b==a

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