Description
Given a string s and 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, and s is 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
Input: s = "abc", t = "ahbgdc", Output: true
Input: s = "axc", t = "ahbgdc", Output: false
Idea
Use two pointers. Start from the beginning of the two strings. When there is a match, advance both; otherwise, advance pointer on t. Returns true if and only if the pointer on s has reach the end.
Solution
class Solution {
public:
bool isSubsequence(string s, string t) {
int m = s.size(), n = t.size();
int p1 = 0, p2 = 0;
while (p1 < m && p2 < n) {
if (s[p1] == t[p2]) {
p1++;
p2++;
} else {
p2++;
}
}
return p1 == m;
}
};
14 / 14 test cases passed.
Runtime: 93 ms
