leetcode刷题系列:392. Is Subsequence
2016-10-31 12:03
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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?
Credits:
Special thanks to @pbrother for adding this problem and creating all test cases.
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Have you met this question in a real interview?
Yes
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?
Credits:
Special thanks to @pbrother for adding this problem and creating all test cases.
Subscribe to see which companies asked this question
Show Tags
Have you met this question in a real interview?
Yes
class Solution { public: bool isSubsequence(string s, string t) { int lens = s.size(); int lent = t.size(); if(lens > lent) return false; int start = 0; int j = 0; for(int i = 0; i < lens; ++i) { for(; j < lent;j++) { if(s[i] == t[j]) { if(j == lent - 1 && i != lens - 1) return false; j++; break; } else { if( j == lent - 1 ) return false; continue; } } } return true; } };
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