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+# Step1
+何も見ずに解く
+```python
+class Solution:
+    def firstUniqChar(self, s: str) -> int:
+        for i, char in enumerate(s):
+            if char not in s[i+1:] and char not in s[:i]:
+                return i
+        return -1
+
+```
+for文で虱潰しに調べた．もっと良いアルゴリズムが有りそう．
+
+# Step2
+他の人の解答を参照
+https://github.com/rinost081/LeetCode/pull/14
+https://github.com/Satorien/LeetCode/pull/15/
+
+- Counterメソッドを使うと文字の出現回数を辞書のように保持できて便利
+- Counterメソッドを持ちいらずに辞書を自分で定義して出現回数を保持するということも出来る
+
+# Step3
+他の人の解答を参考に解き直す
+```python
+class Solution:
+    def firstUniqChar(self, s: str) -> int:
+        char_to_count = Counter(s)
+        for i, char in enumerate(s):
+            if char_to_count[char] == 1:
+                return i
+        return -1
+```
+```python
+class Solution:
+    def firstUniqChar(self, s: str) -> int:
+        char_to_count = defaultdict(int)
+        for char in s:
+            char_to_count[char] += 1
+        for i, char in enumerate(s):
+            if char_to_count[char]==1:
+                return i
+        return -1
+```
+Counterを使うほうが簡潔だし，処理速度も早い．
+
+# Step4
+コメントを受けて再度解き直す
+```python
+class Solution:
+    def firstUniqChar(self, s: str) -> int:
+        char_to_count = defaultdict(int)
+        char_to_index = {}
+        for index, char in enumerate(s):
+            char_to_count[char] += 1
+            if char not in char_to_index:
+                char_to_index[char] = index
+        
+        min_index = len(s)
+        for char, count in char_to_count.items():
+            if count == 1:
+                if char_to_index[char] < min_index:
+                    min_index = char_to_index[char]
+        return min_index if min_index != len(s) else -1
+```
+char_to_indexに各文字の最初のインデックスを保持して，文字列全体に対するfor文の回数を1回で済むように改変した．