按照上面的思路,代码如下: 1classSolution {2intcount = 0;3publicintcountSubstrings(String s) {4char[] array =s.toCharArray();5for(inti = 0 ; i < array.length ; i ++ ) isPalindromic(array,i,i);//以i为中心,长度为奇数6for(inti = 0
方法一参考代码: classSolution{public:intcountSubstrings(strings){intlen = s.size(), res =0;vector<vector<bool>> dp(len,vector<bool>(len,false));for(inti = len -1; i >=0; --i) {for(intj = i; j < len; ++j) { dp[i][j] = (s[i] == s[j]) && (j - i <=2|| d...
Leetcode:647. Palindromic Substrings Leetcode:647. Palindromic Substrings 参考第5题http://www.cnblogs.com/Michael2397/p/8036163.html,一下子就ac掉了,这里有个解释https://discuss.leetcode.com/topic/96884/very-simple-java-solution-with-detail-explanation Oh Those Palindromes 和方案的回文数 分析...
class Solution: def countSubstrings(self, s: str) -> int: # ans 表示 s 中所有回文子串的数量 ans: int = 0 # 枚举回文子串的中心 for i in range(len(s)): # 加上以 s[i] 为中心的回文子串数量 ans += Solution.count(s, i, i) # 加上以 s[i:i+2] 为中心的回文子串数量 ans +=...
/* * @lc app=leetcode id=647 lang=cpp * * [647] Palindromic Substrings */ // @lc code=start class Solution { public: int countSubstrings(string s) { const int N = s.length(); vector<vector<bool>> dp(N, vector<bool>(N, false)); int res = N; for (int i = 0; i < ...
Solution 动态规划,依次求解长度为1,2,3...的子字符串是否为回文,用一个二维数组纪录,同时用一个计数器统计回文子字符串的个数。 Code classSolution{public:intcountSubstrings(strings){if(s.length() ==0)return0;intlen = s.length(); table.resize(len);for(inti =0; i < len; i++) ...
classSolution{int cnt=0;publicintcountSubstrings(String s){if(s==null||s.length()==0)return0;for(int i=0;i<s.length();i++){extendPalidrome(s,i,i);extendPalidrome(s,i,i+1);}returncnt;}privatevoidextendPalidrome(String s,int left,int right){while(left>=0&&right<s.length()&&...
classSolution {publicintcountSubstrings(String s) {if(s ==null|| s.length() == 0){return0; }intcount=0;for(inti = 0; i < s.length(); i++){ count= count + help(s,i,i) + help(s, i, i+1); }returncount; }publicinthelp(String s,inti,intj){intcount = 0;while(i >=...
Code classSolution{publicintcountSubstrings(String s){if(s ==null|| s.length() ==0) {return0; }intn=s.length();boolean[][] f =newboolean[n][n]; f[0][0] =true;char[] c = s.toCharArray();intcount=1;for(inti=1; i < n; i++) { ...
LeetCode 647. Palindromic Substrings(647. 回文子串) 647. Palindromic Substrings Given a string, your task is to count how many palindromic substrings in this string. The substrings with different start indexes or end indexes are counted as different substrings even they consist of same ...