[LeetCode] 392. Is Subsequence

Given a string s and a string t, check if s is subsequence of t.

A subsequence of a string is a new string which is formed from the original string by deleting some (can be none) of the characters without disturbing the relative positions of the remaining characters. (ie, "ace" is a subsequence of "abcde" while "aec" is not).

Follow up:
If there are lots of incoming S, say S1, S2, ... , Sk where k >= 1B, and you want to check one by one to see if T has its subsequence. In this scenario, how would you change your code?

Credits:
Special thanks to @pbrother for adding this problem and creating all test cases.

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 <= 10^4
Both strings consists only of lowercase characters.

判断子序列。题目即是题意。这个题有一个followup，问如果S有很多，怎么处理。我这里给出动态规划的解法。

首先是常规解法，双指针，一个扫描S，一个扫描T。如果碰到同样的字母则两个指针都++，否则只++T的指针。T扫描完毕之后如果S没有到末尾则return false。

时间O(m + n)

空间O(1)

Java实现

 1 class Solution {
 2     public boolean isSubsequence(String s, String t) {
 3         // corner cases 只是s为空
 4         if (s == null || s.length() == 0) {
 5             return true;
 6         }
 7 
 8         // normal case
 9         int i = 0;
10         int j = 0;
11         int sSize = s.length();
12         int tSize = t.length();
13         while (i < sSize && j < tSize) {
14             if (s.charAt(i) == t.charAt(j)) {
15                 i++;
16                 j++;
17             } else {
18                 j++;
19             }
20         }
21         if (i == sSize) {
22             return true;
23         } else {
24             return false;
25         }
26     }
27 }

动态规划。设dp[i][j]为S以i结尾的子串是否是T以j结尾的子串的subsequence。首先corner case是当S为空的时候，dp[0][j] = true。其次开始扫描，如果匹配（s[i] == t[j]），则DP值跟之前一位的DP值相同（dp[i][j] == dp[i - 1][j - 1]）；如果不匹配，则dp[i][j] = dp[i][j - 1]。

时间O(mn)

空间O(mn)

Java实现

 1 class Solution {
 2     public boolean isSubsequence(String s, String t) {
 3         int sLen = s.length();
 4         int tLen = t.length();
 5         if (sLen > tLen) {
 6             return false;
 7         }
 8         if (sLen == 0) {
 9             return true;
10         }
11         boolean[][] dp = new boolean[sLen + 1][tLen + 1];
12         // 初始化
13         for (int j = 0; j < tLen; j++) {
14             dp[0][j] = true;
15         }
16         // dp
17         for (int i = 1; i <= sLen; i++) {
18             for (int j = 1; j <= tLen; j++) {
19                 if (s.charAt(i - 1) == t.charAt(j - 1)) {
20                     dp[i][j] = dp[i - 1][j - 1];
21                 } else {
22                     dp[i][j] = dp[i][j - 1];
23                 }
24             }
25         }
26         return dp[sLen][tLen];
27     }
28 }