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  • Python3解leetcode Isomorphic Strings

    问题描述:

    Given two strings s and t, determine if they are isomorphic.

    Two strings are isomorphic if the characters in s can be replaced to get t.

    All occurrences of a character must be replaced with another character while preserving the order of characters. No two characters may map to the same character but a character may map to itself.

    Example 1:

    Input: s = "egg", t = "add"
    Output: true
    

    Example 2:

    Input: s = "foo", t = "bar"
    Output: false

    Example 3:

    Input: s = "paper", t = "title"
    Output: true

    Note:
    You may assume both and have the same length.

    思路:

    两个字符串对应位置上的字母要具有一一对应的关系,即假如s[1]=s,t[1]=t,则s对应的就是t,如果后面s[i]=s,则t[i]也必须为 t,若t[i]为其他字母,则判定为非同形态。

    这种一一对应关系,首先考虑字典

    代码:

    class Solution:
        def isIsomorphic(self, s: str, t: str) -> bool:
            table = dict()
            for i in range(len(s)):
                if s[i] not in table:
                    table[s[i]] = t[i]
                elif table[s[i]] != t[i]:
                    return False
            return len(table) == len(set(table.values()))

    如果字典中没有相应键值对,那么就在字典中加入该键值对;如果有该键值对,判断一下该元素是否和原存储的键值对相同,不相同则肯定是非同形态,如果相同则是到目前为止为同形态。

    有一种特殊情况,就是不同的键,对应了相同的值,如s='ab',t=‘aa’,出现这种情况代表是非同形态的,通过判断键、值的个数来判断是否出现这种情况。

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  • 原文地址:https://www.cnblogs.com/xiaohua92/p/11158257.html
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