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
andt
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?
class Solution:
def isSubsequence(self, s: str, t: str) -> bool:
i, j, m, n = 0, 0, len(s), len(t)
while i < m and j < n:
if s[i] == t[j]:
i += 1
j += 1
return i == m
class Solution {
public boolean isSubsequence(String s, String t) {
int m = s.length(), n = t.length();
int i = 0, j = 0;
while (i < m && j < n) {
if (s.charAt(i) == t.charAt(j)) {
++i;
}
++j;
}
return i == m;
}
}
class Solution {
public:
bool isSubsequence(string s, string t) {
int m = s.size(), n = t.size();
int i = 0, j = 0;
while (i < m && j < n) {
if (s[i] == t[j]) ++i;
++j;
}
return i == m;
}
};
func isSubsequence(s string, t string) bool {
i, j, m, n := 0, 0, len(s), len(t)
for i < m && j < n {
if s[i] == t[j] {
i++
}
j++
}
return i == m
}
function isSubsequence(s: string, t: string): boolean {
let m = s.length,
n = t.length;
let i = 0;
for (let j = 0; j < n && i < m; ++j) {
if (s.charAt(i) == t.charAt(j)) {
++i;
}
}
return i == m;
}
impl Solution {
pub fn is_subsequence(s: String, t: String) -> bool {
let (s, t) = (s.as_bytes(), t.as_bytes());
let n = t.len();
let mut i = 0;
for &c in s.iter() {
while i < n && t[i] != c {
i += 1;
}
if i == n {
return false;
}
i += 1;
}
true
}
}
bool isSubsequence(char *s, char *t) {
int n = strlen(s);
int i = 0;
for (int j = 0; j < n; j++) {
while (t[i] && t[i] != s[j]) {
i++;
}
if (!t[i]) {
return 0;
}
i++;
}
return 1;
}