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author | jack <[email protected]> | 2024-06-14 13:13:15 -0600 |
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committer | jack <[email protected]> | 2024-06-14 13:13:15 -0600 |
commit | d343a3b18dbaca9297d4f7b2280c4a16f0a5c8aa (patch) | |
tree | 96df922dae2a5073d9a50ea70c7aae6aa37e3ebe /repeated_substring_pattern.cpp | |
download | leetcode-d343a3b18dbaca9297d4f7b2280c4a16f0a5c8aa.tar.gz leetcode-d343a3b18dbaca9297d4f7b2280c4a16f0a5c8aa.zip |
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Diffstat (limited to 'repeated_substring_pattern.cpp')
-rw-r--r-- | repeated_substring_pattern.cpp | 31 |
1 files changed, 31 insertions, 0 deletions
diff --git a/repeated_substring_pattern.cpp b/repeated_substring_pattern.cpp new file mode 100644 index 0000000..4e2d8d8 --- /dev/null +++ b/repeated_substring_pattern.cpp @@ -0,0 +1,31 @@ +// 459. Repeated Substring Pattern +#include "leetcode.h" + +/* + * given a string 's', check if it can be constructed by taking a substring of + * it and appending multiple copies of the substring together. + */ + +class Solution { +public: + bool repeatedSubstringPattern(string s) { + int i = 1, j = 0, n = s.size(); + vector<int> dp(n + 1, 0); + while (i < n) { + if (s[i] == s[j]) + dp[++i] = ++j; + else if (!j) + i++; + else + j = dp[j]; + } + return dp[n] && dp[n] % (n - dp[n]) == 0; + } +}; + +int main() { + Solution obj; + printf("%d\n", obj.repeatedSubstringPattern("abab")); // expect: 1 + printf("%d\n", obj.repeatedSubstringPattern("aba")); // expect: 0 + printf("%d\n", obj.repeatedSubstringPattern("abcabcabcabc")); // expect: 1 +} |