【LeetCode】 392. Is Subsequence
2017-01-07 10:53
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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,
a subsequence of
not).
Example 1:
s =
Return
Example 2:
s =
Return
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?
public class Solution {
public boolean isSubsequence(String s, String t) {
if (s.isEmpty()) {
return true;
}
int p1 = 0, flag = 0;
char[] c1 = s.toCharArray(), c2 = t.toCharArray();
for (int p2 = 0; p2 < c2.length; p2++) {
if (c1[p1] == c2[p2]) {
p1++;
}
if (p1 == c1.length) {
flag = 1;
break;
}
}
return flag == 1;
}
}
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 1:
s =
"abc", t =
"ahbgdc"
Return
true.
Example 2:
s =
"axc", t =
"ahbgdc"
Return
false.
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?
public class Solution {
public boolean isSubsequence(String s, String t) {
if (s.isEmpty()) {
return true;
}
int p1 = 0, flag = 0;
char[] c1 = s.toCharArray(), c2 = t.toCharArray();
for (int p2 = 0; p2 < c2.length; p2++) {
if (c1[p1] == c2[p2]) {
p1++;
}
if (p1 == c1.length) {
flag = 1;
break;
}
}
return flag == 1;
}
}
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