-
Notifications
You must be signed in to change notification settings - Fork 0
Expand file tree
/
Copy path0392_is_subsequence.java
More file actions
76 lines (56 loc) · 2.64 KB
/
0392_is_subsequence.java
File metadata and controls
76 lines (56 loc) · 2.64 KB
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
/*
* https://leetcode-cn.com/problems/is-subsequence/
给定字符串 s 和 t ，判断 s 是否为 t 的子序列。
你可以认为 s 和 t 中仅包含英文小写字母。字符串 t 可能会很长（长度 ~= 500,000），而 s 是个短字符串（长度 <=100）。
字符串的一个子序列是原始字符串删除一些（也可以不删除）字符而不改变剩余字符相对位置形成的新字符串。（例如，"ace"是"abcde"的一个子序列，而"aec"不是）。
示例 1:
s = "abc", t = "ahbgdc"
返回 true.
示例 2:
s = "axc", t = "ahbgdc"
返回 false.
后续挑战: 如果有大量输入的 S，称作S1, S2, ... , Sk 其中 k >= 10亿，你需要依次检查它们是否为 T 的子序列。在这种情况下，你会怎样改变代码？
----------------------------------------------------------------------------------------------------
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 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?
*/
class MySolution {
public boolean isSubsequence(String s, String t) {
if (s.length() > t.length()) return false;
if (s.length() == 0) return true;
char[] sChar = s.toCharArray(), tChar = t.toCharArray();
for (int j = 0; j < t.length(); j++) {
if (t.charAt(j) == sChar[0])
return isSub(sChar, tChar, j);
}
return false;
}
private boolean isSub(char[] s, char[] t, int j) {
int sLen = s.length, tLen = t.length;
if (sLen > tLen - j) return false;
int i = 0;
for (; i < sLen && j < tLen; j++) {
if (sLen - i > tLen - j) return false;
if (s[i] == t[j]) i++;
}
return i == sLen;
}
}
class Solution1 {
public boolean isSubsequence(String s, String t) {
int index = -1;
for (char ch : s.toCharArray()) {
index = t.indexOf(ch, index + 1);
if (index == -1) return false;
}
return true;
}
}