最长回文子串

动态规划

====================================================================

d p [ i ] [ j ] = d p [ i + 1 ] [ j − 1 ]    a n d    s [ i ] = = s [ j ] dp[i][j] = dp[i+1][j-1] \ \ and \ \ s[i]==s[j] dp[i][j]=dp[i+1][j−1]  and  s[i]==s[j]

长串依赖短串的状态。

所以枚举长度的时候从大到小。

const int N = 1010;

class Solution {

public:

string longestPalindrome(string s) {

string ans = “”;

int maxLen = 0, n = s.size();

bool dp[N][N];

for(int len = 1; len <= n ; len ++ ) {

for(int i = 0; i + len -1 < n ; i++ ) {

int j = i + len - 1;

if(len == 1){

dp[i][j] = 1;

}else if(len == 2){

dp[i][j] = s[i] == s[j];

}else{

dp[i][j] = dp[i+1][j-1] && s[i]==s[j];

}

if(dp[i][j] && len > maxLen){

maxLen = len;

ans = s.substr(i,len);

}

}

}

return ans;

}

};

枚举中心点+中心拓展

==========================================================================

class Solution {

public:

pair<int, int> expandAroundCenter(const string& s, int left, int right) {

while (left >= 0 && right < s.size() && s[left] == s[right]) {

–left;

++right;

}

return {left + 1, right - 1};

}

string longestPalindrome(string s) {

int start = 0, end = 0;

for (int i = 0; i < s.size(); ++i) {

auto [left1, right1] = expandAroundCenter(s, i, i);

auto [left2, right2] = expandAroundCenter(s, i, i + 1);

if (right1 - left1 > end - start) {

start = left1;

end = right1;

}

if (right2 - left2 > end - start) {

start = left2;

end = right2;

}

}

return s.substr(start, end - start + 1);

}

};

// 作者：LeetCode-Solution

// 链接：https://leetcode-cn.com/problems/longest-palindromic-substring/solution/zui-chang-hui-wen-zi-chuan-by-leetcode-solution/

// 来源：力扣（LeetCode）

// 著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。

字符串hash+二分

==========================================================================

利用这个过程的单调性，所以可以二分解决，为了快速判断一个串是否是回文串，可以使用字符串hash。实际上是对上面方法的优化。

时间复杂度： O ( n ∗ l o g ( n ) ) O(n*log(n)) O(n∗log(n))

class Solution {

public:

typedef unsigned long long ull;

const static int P = 1e9+7, N = 1010;

ull h1[N] = {0}, h2[N] = {0}, p[N] = {1};

string ans;

int maxlen,n;

string longestPalindrome(string s) {

maxlen = 0, n = s.size();

s = " "+s;

// 求正反字符串hash

for(int i=1;i<=n;i++){

h1[i] = h1[i-1]*P + (s[i]-‘0’+1);

h2[n-i+1] = h2[n-i+2]*P + (s[n-i+1]-‘0’+1);

p[i] = p[i-1]*P;

}

// 奇回文

for(int i = 1; i <= n ; i ++ ){

int l = 0 , r = min(i-1,n-i);

while(l<r){

int mid = l+(r-l+1)/2;

bool f = checkPalindrome(i-mid,i+mid);

if(f){

l = mid;

}else{

r = mid - 1;

}

}

if(2*l+1 > maxlen){

maxlen = 2*l+1;

ans = s.substr(i-l,2*l+1);

}

}

// 偶回文

for(int i = 2; i <= n ; i++ ){

int l = 1 , r = min(i-1,n-i+1);

while(l<r){

int mid = l+(r-l+1)/2;

bool f = checkPalindrome(i-mid,i+mid-1);

if(f){

l = mid;

}else{

r = mid - 1;

}

}

if(2*l > maxlen && checkPalindrome(i-l,i+l-1)){

maxlen = 2*l;

ans = s.substr(i-l,2*l);

}

}

return ans;

}

// 根据字符串hash值判断字符串是否相等

bool checkPalindrome(int l,int r){

if(r>n) return false;

ull hash1 = h1[r] - h1[l-1]*p[r-l+1];

ull hash2 = h2[l] - h2[r+1]*p[r-l+1];

return hash1 == hash2;

}

};

Manacher算法

==========================================================================

class Solution {

public:

string longestPalindrome(string s) {

string str = divide(s);

int strlen = str.size(), maxLen = 0, start = 0;

// manacher算法中最重要的三个量

vector p(strlen,0);

int maxRight = 0, center = 0;

for(int i = 0;i < strlen ; i++) {

int mirror = 2*center - i;

if(i < maxRight){