本文涉及知识点
滚动哈希
二分查找算法合集
LeetCode 1044. 最长重复子串
给你一个字符串 s ,考虑其所有 重复子串 :即 s 的(连续)子串,在 s 中出现 2 次或更多次。这些出现之间可能存在重叠。
返回 任意一个 可能具有最长长度的重复子串。如果 s 不含重复子串,那么答案为 "" 。
示例 1:
输入:s = "banana"
输出:"ana"
示例 2:
输入:s = "abcd"
输出:""
提示:
2 <= s.length <= 3 * 10^4^
s 由小写英文字母组成
二分查找+滚动哈希
令 Check(len) 返回 是否存在长度为len的重复字符串
len1 < len2,如果Check(len2)为true,则Check(len1)一定为true
即 len ∈ \in ∈ [0,len3]为Check(len)为true,len ∈ \in ∈ [len3+1,n] Check(len)为false。
寻找最后一个true,故用左闭右开空间。
Check函数
len = 0 为0,返回true。
用滚动函数计算 s[i...i+len-1]的哈希值, i+ len <= s.length 并将哈希值记录到set中,如果存在重复值,返回true。
时间复杂度 :O(nlogn)
二分查找:O(logn) Check函数O(n)
代码
核心代码
cpp
template<int MOD = 1000000007>
class C1097Int
{
public:
C1097Int(long long llData = 0) :m_iData(llData% MOD)
{
}
C1097Int operator+(const C1097Int& o)const
{
return C1097Int(((long long)m_iData + o.m_iData) % MOD);
}
C1097Int& operator+=(const C1097Int& o)
{
m_iData = ((long long)m_iData + o.m_iData) % MOD;
return *this;
}
C1097Int& operator-=(const C1097Int& o)
{
m_iData = (m_iData + MOD - o.m_iData) % MOD;
return *this;
}
C1097Int operator-(const C1097Int& o)
{
return C1097Int((m_iData + MOD - o.m_iData) % MOD);
}
C1097Int operator*(const C1097Int& o)const
{
return((long long)m_iData * o.m_iData) % MOD;
}
C1097Int& operator*=(const C1097Int& o)
{
m_iData = ((long long)m_iData * o.m_iData) % MOD;
return *this;
}
C1097Int operator/(const C1097Int& o)const
{
return *this * o.PowNegative1();
}
C1097Int& operator/=(const C1097Int& o)
{
*this /= o.PowNegative1();
return *this;
}
bool operator==(const C1097Int& o)const
{
return m_iData == o.m_iData;
}
bool operator<(const C1097Int& o)const
{
return m_iData < o.m_iData;
}
C1097Int pow(long long n)const
{
C1097Int iRet = 1, iCur = *this;
while (n)
{
if (n & 1)
{
iRet *= iCur;
}
iCur *= iCur;
n >>= 1;
}
return iRet;
}
C1097Int PowNegative1()const
{
return pow(MOD - 2);
}
int ToInt()const
{
return m_iData;
}
private:
int m_iData = 0;;
};
//iCodeNum 必须大于等于可能的字符数
template<int MOD = 1000000007>
class CHashStr {
public:
CHashStr(string s, int iCodeNum, int iCodeBegin = 1, char chBegin = 'a') {
m_c = s.length();
m_vP.resize(m_c + 1);
m_vP[0] = 1;
m_vHash.resize(m_c + 1);
for (int i = 0; i < m_c; i++)
{
const int P = iCodeBegin + iCodeNum;
m_vHash[i + 1] = m_vHash[i] * P + s[i] - chBegin + iCodeBegin;
m_vP[i + 1] = m_vP[i] * P;
}
}
//iMinValue将被编码为0,iMaxValue被编码为iMaxValue-iMinValue。
CHashStr(const int* data, int len, int iMinValue = 0, int iMaxValue = 9) {
m_c = len;
m_vP.resize(m_c + 1);
m_vP[0] = 1;
m_vHash.resize(m_c + 1);
const int P = iMaxValue - iMinValue + 1;
for (int i = 0; i < m_c; i++)
{
const int iCurCode = data[i] - iMinValue;
assert((iCurCode >= 0) && (iCurCode < P));
m_vHash[i + 1] = m_vHash[i] * P + iCurCode;
m_vP[i + 1] = m_vP[i] * P;
}
}
//包括left right
int GetHash(int left, int right)
{
return (m_vHash[right + 1] - m_vHash[left] * m_vP[right - left + 1]).ToInt();
}
inline int GetHash(int right)
{
return m_vHash[right + 1].ToInt();
}
int GetHashExincludeRight(int left, int right)
{
return (m_vHash[right] - m_vHash[left] * m_vP[right - left]).ToInt();
}
inline int GetHashExincludeRight(int right)
{
return m_vHash[right].ToInt();
}
int m_c;
vector<C1097Int<MOD>> m_vP;
vector<C1097Int<MOD>> m_vHash;
};
template<int MOD2 = 1000000009>
class C2HashStr
{
public:
C2HashStr(string s) {
m_pHash1 = std::make_unique<CHashStr<>>(s, 26);
m_pHash2 = std::make_unique < CHashStr<MOD2>>(s, 27, 0);
}
C2HashStr(const int* data, int len, int iMinValue = 0, int iMaxValue = 9)
{
m_pHash1 = std::make_unique<CHashStr<>>(data, len, iMinValue, iMaxValue);
m_pHash2 = std::make_unique < CHashStr<MOD2>>(data, len, iMinValue, iMaxValue);
}
//包括left right
long long GetHash(int left, int right)
{
return (long long)m_pHash1->GetHash(left, right) * (MOD2 + 1) + m_pHash2->GetHash(left, right);
}
long long GetHash(int right)
{
return (long long)m_pHash1->GetHash(right) * (MOD2 + 1) + m_pHash2->GetHash(right);
}
//包括Left,不包括Right
long long GetHashExincludeRight(int left, int right)
{
return (long long)m_pHash1->GetHashExincludeRight(left, right) * (MOD2 + 1) + m_pHash2->GetHashExincludeRight(left, right);
}
long long GetHashExincludeRight(int right)
{
return (long long)m_pHash1->GetHashExincludeRight(right) * (MOD2 + 1) + m_pHash2->GetHashExincludeRight(right);
}
private:
std::unique_ptr<CHashStr<>> m_pHash1;
std::unique_ptr<CHashStr<MOD2>> m_pHash2;
};
namespace NBinarySearch
{
template<class INDEX_TYPE, class _Pr>
INDEX_TYPE FindFrist(INDEX_TYPE left, INDEX_TYPE rightInclue, _Pr pr)
{
while (rightInclue - left > 1)
{
const auto mid = left + (rightInclue - left) / 2;
if (pr(mid))
{
rightInclue = mid;
}
else
{
left = mid;
}
}
return rightInclue;
}
template<class INDEX_TYPE, class _Pr>
INDEX_TYPE FindEnd(INDEX_TYPE leftInclude, INDEX_TYPE right, _Pr pr)
{
while (right - leftInclude > 1)
{
const auto mid = leftInclude + (right - leftInclude) / 2;
if (pr(mid))
{
leftInclude = mid;
}
else
{
right = mid;
}
}
return leftInclude;
}
}
class Solution {
public:
string longestDupSubstring(string s) {
string ret;
C2HashStr<> dh(s);
auto Check = [&](int len) {
if (0 == len) { ret = ""; return true; }
unordered_set<long long> setHas;
for (int i = 0; i + len <= s.length(); i++) {
auto cur = dh.GetHashExincludeRight(i, i + len);
if (setHas.count(cur)) {
ret = s.substr(i, len);
return true;
}
setHas.emplace(cur);
}
return false;
};
NBinarySearch::FindEnd(0, (int)s.length() + 1, Check);
return ret;
}
};
单元测试
cpp
template<class T1,class T2>
void AssertEx(const T1& t1, const T2& t2)
{
Assert::AreEqual(t1 , t2);
}
template<class T>
void AssertEx(const vector<T>& v1, const vector<T>& v2)
{
Assert::AreEqual(v1.size(), v2.size());
for (int i = 0; i < v1.size(); i++)
{
Assert::AreEqual(v1[i], v2[i]);
}
}
template<class T>
void AssertV2(vector<vector<T>> vv1, vector<vector<T>> vv2)
{
sort(vv1.begin(), vv1.end());
sort(vv2.begin(), vv2.end());
Assert::AreEqual(vv1.size(), vv2.size());
for (int i = 0; i < vv1.size(); i++)
{
AssertEx(vv1[i], vv2[i]);
}
}
namespace UnitTest
{
string s;
TEST_CLASS(UnitTest)
{
public:
TEST_METHOD(TestMethod1)
{
s = "banana";
auto res = Solution().longestDupSubstring(s);
AssertEx(string("ana"), res);
}
TEST_METHOD(TestMethod2)
{
s = "abcd";
auto res = Solution().longestDupSubstring(s);
AssertEx(string(""), res);
}
TEST_METHOD(TestMethod3)
{
s = "aa";
auto res = Solution().longestDupSubstring(s);
AssertEx(string("a"), res);
}
};
}
扩展阅读
视频课程
先学简单的课程,请移步CSDN学院,听白银讲师(也就是鄙人)的讲解。
https://edu.csdn.net/course/detail/38771
如何你想快速形成战斗了,为老板分忧,请学习C#入职培训、C++入职培训等课程
https://edu.csdn.net/lecturer/6176
相关推荐
我想对大家说的话 |
---|
《喜缺全书算法册》以原理、正确性证明、总结为主。 |
按类别查阅鄙人的算法文章,请点击《算法与数据汇总》。 |
有效学习:明确的目标 及时的反馈 拉伸区(难度合适) 专注 |
闻缺陷则喜(喜缺)是一个美好的愿望,早发现问题,早修改问题,给老板节约钱。 |
子墨子言之:事无终始,无务多业。也就是我们常说的专业的人做专业的事。 |
如果程序是一条龙,那算法就是他的是睛 |
测试环境
操作系统:win7 开发环境: VS2019 C++17
或者 操作系统:win10 开发环境: VS2022 C++17
如无特殊说明,本算法用**C++**实现。