[LeetCode][Python]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.

解题思路：

对于这种在两个字符串中取不同的，可以考虑推导表达式，但是没考虑到s和t的组成字母是同一个元素的情况。

第一思路还是从t中移走所有s中的元素，剩下的即为所求元素。这个可以接受，但是只击败了5.48%Python的答案。可见很不优。

考虑到两个字符串里面只有一个不同，一开始想到使用异或处理，不过下面怎么继续有点不是很清楚。后来看了别人的答案，原来可以和ord以及chr函数搭配使用。ord()函数是chr()函数（对于8位的ASCII字符串）或unichr()函数（对于Unicode对象）的配对函数，它以一个字符（长度为1的字符串）作为参数，返回对应的ASCII数值，或者Unicode数值。可以分别对s和t进行操作，也可以对s+t进行操作。

#!/usr/bin/env python
# -*- coding: UTF-8 -*-
class Solution(object):
    def findTheDifference(self, s, t):
        """
        :type s: str
        :type t: str
        :rtype: str
        """
        l1, l2 = list(s), list(t)
        for e in l1:
            l2.remove(e)
        return "".join(l2)
        # return "".join([ele for ele in t if ele not in s])

    def findTheDifference2(self, s, t):
        code = 0
        for ch in s:
            code ^= ord(ch)
        for ch in t:
            code ^= ord(ch)
        return chr(code)

    def findTheDifference3(self, s, t):
        ans = 0
        for c in s+t:
            ans ^= ord(c)
        return chr(ans)


if __name__ == '__main__':
    sol = Solution()
    s = "abcd"
    t = "abcde"
    print sol.findTheDifference(s, t)
    print sol.findTheDifference2(s, t)