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Merge pull request neetcode-gh#2025 from andrewmustea/c_0392-is-subse…
…quence_use_pointers c: 0392-is-subsequence: use pointers instead of recursion
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/* | ||
Given two strings s and t, return true if s is a subsequence of t, or false otherwise. | ||
/** | ||
* Given two strings s and t, return true if s is a subsequence of t, or false | ||
* otherwise. A subsequence of a string is a new string that is formed from the | ||
* original string by deleting some (can be none) of the characters without | ||
* disturbing the relative positions of the remaining characters. (i.e., "ace" | ||
* is a subsequence of "abcde" while "aec" is not). | ||
* | ||
* Example 1: | ||
Space: O(1) | ||
Time: O(n) | ||
*/ | ||
* Input: s = "abc", t = "ahbgdc" | ||
* Output: true | ||
* | ||
* Example 2: | ||
* | ||
* Input: s = "axc", t = "ahbgdc" | ||
* Output: false | ||
* | ||
* Constraints: | ||
* | ||
* 0 <= s.length <= 100 | ||
* 0 <= t.length <= 104 | ||
* s and t consist only of lowercase English letters. | ||
* | ||
* Space: O(1) | ||
* Time: O(n) | ||
*/ | ||
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bool isSubsequence(char * s, char * t){ | ||
if (s[0]=='\0'){ | ||
return true; | ||
} else if (t[0]=='\0') { | ||
return false; | ||
} else if (t[0]==s[0]) { | ||
return isSubsequence(s+1, t+1); | ||
} else { | ||
return isSubsequence(s, t+1); | ||
char *s_char = s; | ||
char *t_char = t; | ||
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while (*s_char != 0) { | ||
if (*t_char == 0) { | ||
return false; | ||
} | ||
else if (*s_char == *t_char) { | ||
++s_char; | ||
} | ||
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++t_char; | ||
} | ||
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return true; | ||
} |