392 Is Subsequence
392. Is Subsequence
1. Question
Given a stringsand a string t, check if s is subsequence of t.
You may assume that there is only lower case English letters in both s and t. t is potentially a very long (length ~= 500,000) string, andsis a short string (<=100).
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).
Example 1:
s="abc"
, t="ahbgdc"
Returntrue
.
Example 2:
s="axc"
, t="ahbgdc"
Returnfalse
.
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?
2. Implementation
(1) DP
思路: dp[i][j] 表示s[0...i - 1]是不是t[0...j - 1]的subsequence
class Solution {
public boolean isSubsequence(String s, String t) {
int m = s.length(), n = t.length();
boolean[][] dp = new boolean[s.length() + 1][t.length() + 1];
for (int i = 0; i <= m; i++) {
for (int j = 0; j <= n; j++) {
if (i == 0) {
dp[i][j] = true;
}
else if (j == 0) {
dp[i][j] = false;
}
else if (s.charAt(i - 1) == t.charAt(j - 1)) {
dp[i][j] = dp[i - 1][j - 1];
}
else {
dp[i][j] = dp[i][j - 1];
}
}
}
return dp[m][n];
}
}
3. Time & Space Complexity
DP: 时间复杂度O(mn), 空间复杂度O(mn)
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