387. First Unique Character in a String

README.md

@@ -1,6 +1,6 @@
 # LeetCode
 
-## 問題: 
+## 問題: [387. First Unique Character in a String](https://leetcode.com/problems/first-unique-character-in-a-string/description/)
 
 ## 前提
 

step1.md

@@ -1,7 +1,29 @@
 # Step 1
 
+- setとdictを用意する
+- 各characterをsetに追加、初見ならdictにインデックスと一緒に追加
+- characterが再び現れたらdictから削除
+- ループが一周したら、dictでループし最小のインデックスを返す
+
 ```python
+class Solution:
+    def firstUniqChar(self, s: str) -> int:
+        chars = set()
+        unique_char_to_idx = dict()
+
+        for idx, char in enumerate(s):
+            if char in chars:
+                unique_char_to_idx.pop(char, None)
+            else:
+                chars.add(char)
+                unique_char_to_idx[char] = idx
+
+        if len(unique_char_to_idx) == 0:
+            return -1
+
+        return min(unique_char_to_idx.values())
 ```
-時間計算量:
 
-空間計算量:
+時間計算量: O(n)
+
+空間計算量: O(n)

step2.md

@@ -1,4 +1,19 @@
 # Step 2
 
+- hashを2つ宣言する必要なかった
+
 ```python
+class Solution:
+    def firstUniqChar(self, s: str) -> int:
+        character_to_idx = dict()
+        n = len(s)
+
+        for idx, character in enumerate(s):
+            if character in character_to_idx:
+                character_to_idx[character] = n
+            else:
+                character_to_idx[character] = idx
+
+        res = min(character_to_idx.values())
+        return -1 if res == n else res
 ```

step3.md

@@ -1,9 +1,23 @@
 # Step 3
 
 ```python
+class Solution:
+    def firstUniqChar(self, s: str) -> int:
+        character_to_idx = dict()
+        n = len(s)
+
+        for idx, character in enumerate(s):
+            if character in character_to_idx:
+                character_to_idx[character] = n
+            else:
+                character_to_idx[character] = idx
+
+        res = min(character_to_idx.values())
+        return -1 if res == n else res
 ```
-1回目: 分　秒
 
-2回目: 分　秒
+1回目: 1分　53秒
+
+2回目: 1分　54秒
 
-3回目: 分　秒
+3回目: 1分　45秒