【LeetCode】2351. First Letter to Appear Twice 解答・解説【Python】

問題
1. 原文
2. 内容
解答
1. 解答1：Python,　dict
2. 解答2：Python, set
メモ・参考・感想

問題

原文

Given a string s consisting of lowercase English letters, return the first letter to appear twice.

Note:

A letter a appears twice before another letter b if the second occurrence of a is before the second occurrence of b.

s will contain at least one letter that appears twice.

Example 1:

Input: s = "abccbaacz" Output: "c" Explanation: The letter 'a' appears on the indexes 0, 5 and 6. The letter 'b' appears on the indexes 1 and 4. The letter 'c' appears on the indexes 2, 3 and 7. The letter 'z' appears on the index 8. The letter 'c' is the first letter to appear twice, because out of all the letters the index of its second occurrence is the smallest.

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Input: s = "abccbaacz"
Output: "c"
Explanation:
The letter 'a' appears on the indexes 0, 5 and 6.
The letter 'b' appears on the indexes 1 and 4.
The letter 'c' appears on the indexes 2, 3 and 7.
The letter 'z' appears on the index 8.
The letter 'c' is the first letter to appear twice, because out of all the letters the index of its second occurrence is the smallest.

Example 2:

Input: s = "abcdd" Output: "d" Explanation: The only letter that appears twice is 'd' so we return 'd'.

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Input: s = "abcdd"
Output: "d"
Explanation:
The only letter that appears twice is 'd' so we return 'd'.

Constraints:

2 <= s.length <= 100

s consists of lowercase English letters.

s has at least one repeated letter.

内容

英子文字で構成される文字列sが与えられるので、最初に2回出現する文字を返してください。

・もし、aの2回目の出現がbの2回目の出現よりも前に発生した場合、aはbが現れる前に2回出現したとします。

・文字列sは少なくとも1つの2回出現する文字を持ちます。

※正しくない可能性があります。

解答

解答1：Python,　dict

class Solution:
    def repeatedCharacter(self, s: str) -> str:
        count_dict = {}
        for i in s:
            if i in count_dict.keys():
                count_dict[i] += 1
                if count_dict[i] == 2:
                    return i
            else:
                count_dict[i] = 1

class Solution:

def repeatedCharacter(self, s: str) -> str:

count_dict = {}

for i in s:

if i in count_dict.keys():

count_dict[i] += 1

if count_dict[i] == 2:

return i

else:

count_dict[i] = 1

初めて出現したときはcount_dictにその文字と1を追加。

2回目に出現したときはcount_dict内の値を1増やす。

その値が2になっていれば関数の戻り値とする。

解答2：Python, set

class Solution:
    def repeatedCharacter(self, s: str) -> str:
        letters = set()
        for i in s:
            if i in letters:
                return i
            else:
                letters.add(i)

class Solution:

def repeatedCharacter(self, s: str) -> str:

letters = set()

for i in s:

if i in letters:

return i

else:

letters.add(i)

set型変数lettersを用意。

sの文字列がlettersに既にあれば戻り値とする。

sの文字列がlettersになければlettersに加える。

メモ・参考・感想

前：300. Longest Increasing Subsequence

次：107. Binary Tree Level Order Traversal II

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問題

原文

内容

解答

解答1：Python, dict

解答2：Python, set

メモ・参考・感想

コメント

解答1：Python,　dict