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
.
这道题算比较简单的一种,我们可以用两个指针分别指向字符串s和t,然后如果字符相等,则i和j自增1,反之只有j自增1,最后看i是否等于s的长度,等于说明s已经遍历完了,而且字符都有在t中出现过,参见代码如下:
解法一:
class Solution { public: bool isSubsequence(string s, string t) { if (s.empty()) return true; int i = 0, j = 0; while (i < s.size() && j < t.size()) { if (s[i] == t[j]) { ++i; ++j; } else { ++j; } } return i == s.size(); } };
下面这种写法稍稍简洁了一些,但是思路并没有什么不同,参见代码如下:
解法二:
class Solution { public: bool isSubsequence(string s, string t) { if (s.empty()) return true; int i = 0, j = 0; while (i < s.size() && j < t.size()) { if (s[i] == t[j]) ++i; ++j; } return i == s.size(); } };
本文转自博客园Grandyang的博客,原文链接:是子序列[LeetCode] Is Subsequence ,如需转载请自行联系原博主。