forked from shuboc/LeetCode-2
-
Notifications
You must be signed in to change notification settings - Fork 0
/
shortest-palindrome.cpp
104 lines (96 loc) · 3.05 KB
/
shortest-palindrome.cpp
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
// Time: O(n)
// Space: O(n)
// KMP Algorithm
class Solution {
public:
string shortestPalindrome(string s) {
if (s.empty()) {
return s;
}
string rev_s(s.crbegin(), s.crend());
// Assume s is (Palindrome)abc,
// A would be (Palindrome)abccba(Palindrome).
string A = s + rev_s;
auto prefix = getPrefix(A);
// The index prefix.back() of A would be:
// (Palindrome)abccba(Palindrome)
// ^
// The index prefix.back() + 1 of s would be:
// (Palindrome)abc
// ^
// Get non palindrome part of s.
int i = prefix.back();
while (i >= s.length()) {
i = prefix[i];
}
string non_palindrome = s.substr(i + 1);
reverse(non_palindrome.begin(), non_palindrome.end());
return non_palindrome + s; // cba(Palindrome)abc.
}
private:
vector<int> getPrefix(const string& pattern) {
vector<int> prefix(pattern.length(), -1);
int j = -1;
for (int i = 1; i < pattern.length(); ++i) {
while (j > -1 && pattern[j + 1] != pattern[i]) {
j = prefix[j];
}
if (pattern[j + 1] == pattern[i]) {
++j;
}
prefix[i] = j;
}
return prefix;
}
};
// Time: O(n)
// Space: O(n)
// Manacher's Algorithm
class Solution2 {
public:
string shortestPalindrome(string s) {
string T = preProcess(s);
int n = T.length();
vector<int> P(n);
int C = 0, R = 0;
for (int i = 1; i < n - 1; ++i) {
int i_mirror = 2 * C - i; // equals to i' = C - (i-C)
P[i] = (R > i) ? min(R - i, P[i_mirror]) : 0;
// Attempt to expand palindrome centered at i
while (T[i + 1 + P[i]] == T[i - 1 - P[i]]) {
++P[i];
}
// If palindrome centered at i expand past R,
// adjust center based on expanded palindrome.
if (i + P[i] > R) {
C = i;
R = i + P[i];
}
}
// Find the max len of palindrome which starts with the first char of s.
int max_len = 0;
for (int i = 1; i < n - 1; ++i) {
if (i - P[i] == 1) {
max_len = P[i];
}
}
// Assume s is (Palindrome)abc.
string ans = s.substr(max_len); // abc.
reverse(ans.begin(), ans.end()); // cba.
ans.append(s); // cba(Palindrome)abc.
return ans;
}
private:
string preProcess(string s) {
int n = s.length();
if (n == 0) {
return "^$";
}
string ret = "^";
for (int i = 0; i < n; ++i) {
ret += "#" + s.substr(i, 1);
}
ret += "#$";
return ret;
}
};