Create 387. First Unique Character in a String.md

387. First Unique Character in a String/387. First Unique Character in a String.md

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+https://leetcode.com/problems/first-unique-character-in-a-string/description/
+
+## STEP 1
+カウンターで頻度を取っておいて、次は先頭から頻度が1のものを探す。
+```python
+from collections import Counter
+class Solution:
+    def firstUniqChar(self, s: str) -> int:
+        counter = Counter(s)
+        for index, char in enumerate(s):
+            if counter[char] == 1:
+                return index
+        return -1
+```
+
+## STEP 2
+ Linked Hashmapを使っても解けるみたい。順序を保持するhashmapに今の所1回しか登場していない文字と場所を記録し、すでに出現した文字をhashsetに記録する。2回目が現れたらhashmapから消して、最後にhashmapの中で一番最近のものを取得する。
+ https://docs.google.com/document/d/11HV35ADPo9QxJOpJQ24FcZvtvioli770WWdZZDaLOfg/edit?tab=t.0#heading=h.hdbcrwo3urck
+ ```python
+  class Solution:
+    def firstUniqChar(self, s: str) -> int:
+        unique_char_to_index = {}
+        seen_chars = set()
+        for i, c in enumerate(s):
+            if c in unique_char_to_index:
+                del unique_char_to_index[c]
+                continue
+            if c in seen_chars:
+                continue
+            seen_chars.add(c)
+            unique_char_to_index[c] = i
+        if not unique_char_to_index:
+            return -1
+        return next(iter(unique_char_to_index.values()))
+ ```
+## STEP 3
+STEP2の方針で練習。
+```python
+class Solution:
+    def firstUniqChar(self, s: str) -> int:
+        seen = set()
+        unique_char_to_index = {}
+        for i, c in enumerate(s):
+            if c in unique_char_to_index:
+                del unique_char_to_index[c]
+                continue
+            if c in seen:
+                continue
+            seen.add(c)
+            unique_char_to_index[c] = i
+        if not unique_char_to_index:
+            return -1
+        return next(iter(unique_char_to_index.values()))
+```