387. First Unique Character in a String

lc387.md

@@ -0,0 +1,69 @@
+# step1
+思考ログ
+- 2つのpointerを使って走査すれば最悪計算時間はO(n^2)かかるが単純だと思った.
+
+```python
+class Solution:
+    def firstUniqChar(self, s: str) -> int:
+        seen = set()
+        for i in range(len(s)):
+            if s[i] in seen:
+                continue
+            j = i + 1
+            is_repeat = False
+            while(j < len(s)):
+                if s[j] == s[i]:
+                    is_repeat = True
+                j += 1
+
+            if is_repeat is False:
+                return i
+            seen.add(s[i])
+
+        return -1
+```
+
+# step2
+- step1をchatgptに投げた結果, if s[j] == s[i]:のあとにbreakを入れるべきという指摘を受けた. その通りだと思った.
+参考にした他の方のPR
+- https://github.com/potrue/leetcode/pull/15/files?short_path=5e4287a#diff-5e4287a02e3f11dbbbe1217cb91b9370bcf7798cdde8ce6e7c1729817e53c0fc
+- 前回同様,私のstep1での実装をライブラリ(Counterとfind)で簡潔に解いている実装だった.明らかにこちらの方が読みやすい.  
+```python
+class Solution:
+    def firstUniqChar(self, s: str) -> int:
+        str_to_count = Counter(s)
+        for character in s:
+            if str_to_count[character] == 1:
+                return s.find(character)
+
+        return -1
+```
+- enumerateで場所を特定する方法
+```python
+class Solution:
+    def firstUniqChar(self, s: str) -> int:
+        str_to_count = Counter(s)
+        for i, character in enumerate(s):
+            if str_to_count[character] == 1:
+                return i
+
+        return -1
+```
+- https://github.com/nktr-cp/leetcode/pull/16/files?short_path=0c860cd#diff-0c860cd754249868513e4f9054206317fa33d0f548fc3896ac2b3e11822fd852
+- c++での実装、ただ眺めていた
+
+- https://github.com/mura0086/arai60/pull/19/files#diff-5ec7c3c87171edf4d61e9eb79fd926cafa27caf068da7474222897c8e9e7ab96
+- [lru_cache](https://docs.python.org/ja/3.13/library/functools.html#functools.lru_cache)(Least Recently Used cache)という概念を初めて知った. 
+
+# step3
+```
+class Solution:
+    def firstUniqChar(self, s: str) -> int:
+        char_to_count = Counter(s)
+
+        for i, character in enumerate(s):
+            if char_to_count[character] == 1:
+                return i
+
+        return -1
+```