Create Is Subsequence.md

Is Subsequence/Is Subsequence.md

@@ -0,0 +1,86 @@
+### Is Subsequence
+
+https://leetcode.com/problems/is-subsequence
+
+#### step1
+```python3
+class Solution:
+    def isSubsequence(self, s: str, t: str) -> bool:
+        # 先頭からみて同じものがあれば消費する。最後までindexを回せたら終わり
+        # 終了判定を挟む必要があり、そこをわかりやすくかけるかどうかでindexの扱いを変えたい
+        if len(s) == 0:
+            return True
+        seen = 0
+        for c in t:
+            if s[seen] == c:
+                seen += 1
+            if seen == len(s):
+                return True
+        return False
+```
+
+#### step2
+[Arai60/392. Is Subsequence.md at 0aac26e68fa6b8f516ff494da34578b8cc1cefc2 · olsen-blue/Arai60 · GitHub](https://github.com/olsen-blue/Arai60/blob/0aac26e68fa6b8f516ff494da34578b8cc1cefc2/392.%20Is%20Subsequence.md#%E8%A7%A3%E6%B3%952%E4%BB%95%E4%BA%8B%E3%81%AE%E5%BC%95%E3%81%8D%E7%B6%99%E3%81%8E%E3%83%88%E3%83%83%E3%83%97%E3%83%80%E3%82%A6%E3%83%B3%E5%86%8D%E5%B8%B0dpac)
+再帰で書くループ
+やっていることはstep1と同じだが、再帰苦手なので、こっちでも実装する
+
+```python3
+class Solution:
+    def isSubsequence(self, s: str, t: str) -> bool:
+        def traverse(s_index, t_index) -> bool:
+            if s_index == len(s):
+                return True
+            if t_index == len(t):
+                return False
+
+            if s[s_index] == t[t_index]:
+                return traverse(s_index + 1, t_index + 1)
+            else:
+                return traverse(s_index, t_index + 1)
+        return traverse(0, 0)
+```
+return traverse(s_index + 1, t_index + 1)のreturnをつけ忘れて1敗
+
+再帰をwhile True:で書き直す方法も真似る
+```python3
+class Solution:
+    def isSubsequence(self, s: str, t: str) -> bool:
+        s_index = 0
+        t_index = 0
+
+        while True:
+            if s_index == len(s):
+                return True
+            if t_index == len(t):
+                return False
+
+            if s[s_index] == t[t_index]:
+                s_index += 1
+                t_index += 1
+            else:
+                t_index += 1
+```
+
+
+[392. Is Subsequence by fhiyo · Pull Request #55 · fhiyo/leetcode · GitHub](https://github.com/fhiyo/leetcode/pull/55/files#diff-a6c7d5ff748fd033529b0b0a550ed2aa570e18edc3e2c61da5094aec0e23a91eR45-R46)
+- 二分探索法
+- s<<tでかつ、sが違う処理が入るときにtを前処理して置きたいという気持ちから
+- 計算量はO(len(s)loglen(t))でO(len(t))よりは短くなる状況はありそう。
+
+#### step3
+```python3
+class Solution:
+    def isSubsequence(self, s: str, t: str) -> bool:
+        if s == "":
+            return True
+
+        seen = 0
+        for char_t in t:
+            if s[seen] == char_t:
+                seen += 1
+            if seen == len(s):
+                return True
+        return False
+```
+
+1:14, 57, 56sec