Given a string s, find the longest palindromic substring in s. You may assume that the maximum length of s is 1000.
Example:
Input: "babad"
Output: "bab"
Note: "aba" is also a valid answer.
Example:
Input: "cbbd"
Output: "bb"
這道題需要使用動態規劃,通過維護一個二維數組dp,其中dp[i][j]表示字符串區間[i, j]是否爲迴文串,當i = j時,只有一個字符,肯定是迴文串,如果i = j + 1,說明是相鄰字符,此時需要判斷s[i]是否等於s[j],如果i和j不相鄰,即i - j >= 2時,除了判斷s[i]和s[j]相等之外,dp[j + 1][i - 1]若爲真,就是迴文串,通過以上分析,可以寫出遞推式如下:
dp[i, j] = 1 if i == j
= s[i] == s[j] if j = i + 1
= s[i] == s[j] && dp[i + 1][j - 1] if j > i + 1
實現的代碼如下:
public:
string longestPalindrome(string s) {
int dp[s.size()][s.size()] = {0}, left = 0, right = 0, len = 0;
for (int i = 0; i < s.size(); ++i) {
for (int j = 0; j < i; ++j) {
dp[j][i] = (s[i] == s[j] && (i - j < 2 || dp[j + 1][i - 1]));
if (dp[j][i] && len < i - j + 1) {
len = i - j + 1;
left = j;
right = i;
}
}
dp[i][i] = 1;
}
return s.substr(left, right - left + 1);
}
};
但是提交之後發現效率並不是特別的高,有查了一些資料原來還有一種算法可以將複雜度遠遠降低,馬拉車算法
用該算法解決該問題的代碼如下:
class Solution {
public:
string longestPalindrome(string s) {
string t ="$#";
for (int i = 0; i < s.size(); ++i) {
t += s[i];
t += '#';
}
int p[t.size()] = {0}, id = 0, mx = 0, resId = 0, resMx = 0;
for (int i = 0; i < t.size(); ++i) {
p[i] = mx > i ? min(p[2 * id - i], mx - i) : 1;
while (t[i + p[i]] == t[i - p[i]]) ++p[i];
if (mx < i + p[i]) {
mx = i + p[i];
id = i;
}
if (resMx < p[i]) {
resMx = p[i];
resId = i;
}
}
return s.substr((resId - resMx) / 2, resMx - 1);
}
};
果然算法的複雜度提升了。