feat: solve No.392

Author: BaffinLee

301-400/392. Is Subsequence.md

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+# 392. Is Subsequence
+
+- Difficulty: Easy.
+- Related Topics: Two Pointers, String, Dynamic Programming.
+- Similar Questions: Number of Matching Subsequences, Shortest Way to Form String, Append Characters to String to Make Subsequence, Make String a Subsequence Using Cyclic Increments.
+
+## Problem
+
+Given two strings `s` and `t`, return `true`** if **`s`** is a **subsequence** of **`t`**, or **`false`** otherwise**.
+
+A **subsequence** of a string is a new string that is formed from the original string by deleting some (can be none) of the characters without disturbing the relative positions of the remaining characters. (i.e., `"ace"` is a subsequence of `"abcde"` while `"aec"` is not).
+
+ 
+Example 1:
+```
+Input: s = "abc", t = "ahbgdc"
+Output: true
+```Example 2:
+```
+Input: s = "axc", t = "ahbgdc"
+Output: false
+```
+ 
+**Constraints:**
+
+
+	
+- `0 <= s.length <= 100`
+	
+- `0 <= t.length <= 104`
+	
+- `s` and `t` consist only of lowercase English letters.
+
+
+ 
+**Follow up:** Suppose there are lots of incoming `s`, say `s1, s2, ..., sk` where `k >= 109`, and you want to check one by one to see if `t` has its subsequence. In this scenario, how would you change your code?
+
+## Solution
+
+```javascript
+/**
+ * @param {string} s
+ * @param {string} t
+ * @return {boolean}
+ */
+var isSubsequence = function(s, t) {
+    var j = 0;
+    for (var i = 0; i < s.length; i++) {
+        while (s[i] !== t[j] && j < t.length - 1) j++;
+        if (s[i] != t[j]) return false;
+        j++;
+    }
+    return true;
+};
+```
+
+**Explain:**
+
+nope.
+
+**Complexity:**
+
+* Time complexity : O(n).
+* Space complexity : O(1).