leetcode 389. Find the Difference

找出字符串中新增的字符

最新推荐文章于 2025-02-20 00:20:29 发布

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最新推荐文章于 2025-02-20 00:20:29 发布

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本文介绍了一种算法，用于找出一个字符串相对于另一个字符串增加的一个字符。通过三种方法实现：利用字典计数、排序比较及位操作进行高效查找。

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.

class Solution(object):
    def findTheDifference(self, s, t):
        """
        :type s: str
        :type t: str
        :rtype: str
        """
        """
        s = "abcd" t = "abcde" =>e
        s = "abcd" t = "abcdee" =>ee
        s = "abcd" t = "abcdfe" =>ef?fe?
        s = "a" t = "" =>a
        s = "ab" t = "" =>ab?ba?
        # use dict to map t
        # sub the letter of t in s
        """
        letter_cnt = {}
        for c in t:
            if c not in letter_cnt:
                letter_cnt[c] = 0
            letter_cnt[c] += 1
        for c in s:
            if c in letter_cnt:
                letter_cnt[c] -= 1
                assert letter_cnt[c] >= 0
        return "".join(c*letter_cnt[c] for c in letter_cnt)

用排序：

class Solution(object):
    def findTheDifference(self, s, t):
        """
        :type s: str
        :type t: str
        :rtype: str
        """
        # abcd, abcde
        # afjz, abcfjz
        # afjx, abcfjxyz
        s = sorted(s)
        t = sorted(t)
        ans = ""
        i = j = 0
        while i< len(s):
            if s[i] != t[j]:
                ans += t[j]
                j += 1
            else:
                i += 1
                j += 1
        return ans+"".join(t[j:])

如果仅仅有一个char不一样则可以使用xor：

class Solution(object):
    def findTheDifference(self, s, t):
        """
        :type s: str
        :type t: str
        :rtype: str
        """
        # if just only one char is different
        ans = 0
        for c in s:
            ans ^= ord(c)
        for c in t:
            ans ^= ord(c)
        return chr(ans)

class Solution(object):
    def findTheDifference(self, s, t):
        """
        :type s: str
        :type t: str
        :rtype: str
        """
        # if just only one char is different        
        return chr(sum(ord(i) for i in t)-sum(ord(i) for i in s))