代码随想录算法训练营第五十七天|动态规划part17|647. 回文子串 ● 516.最长回文子序列● 动态规划总结篇

647. 回文子串  Palindromic Substrings - LeetCode

dp[i][j]  [i,j]范围内的子串是否是回文子串

 1.i == j

2.i j相差1

3. j - 1 > i dp[i - 1][j - 1]

if (s[i] ==s[j])

        if （j - i <= 1) dp[i][j] = true; res++;

        else if (dp[i - 1][j - 1] == true) dp[i][j] = true; res++;

初始化：dp[i][j]

dp[i][j] = false

遍历顺序： 从低往上，从左往右

for (int i = size() - 1; i >= 0; i--)

        for (int j = i; j < s.length(); j++)

class Solution {
    public int countSubstrings(String s) {
        int n = s.length();
        boolean[][] dp = new boolean[n][n];
       // dp[0][0] = false;
        int res = 0;

        for (int i = s.length() - 1; i >= 0; i--) {
            for (int j = i; j < s.length(); j++) {
                if (s.charAt(i) == s.charAt(j)) {
                    if (j - i <= 1) {
                        dp[i][j] = true;
                        res++;
                    } else if (dp[i + 1][j - 1]) {
                        dp[i][j] = true;
                        res++;
                    }

                }
            }
        }
        return res;
    }
}

●  516.最长回文子序列 Longest Palindromic Subsequence - LeetCode

 dp[i][j] [i,j]的回文子串并列长度为

if (s[i] == s[j]) dp[i][j] = dp[i + 1][j - 1] + 2//左右都加了一个元素

else  dp[i][j] = max(dp[i][j - 1], dp[i + 1][j])

初始化： dp[i][i] = 1;

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

        dp[i][i] = 1;

遍历顺序：

for (i < s.length() - 1; i >= 0; i--)

        for (int j = i + 1; j < s.length(); j++)

class Solution {
    public int longestPalindromeSubseq(String s) {
        int[][] dp = new int[s.length()][s.length()];
        for (int i = 0; i < s.length(); i++) {
            dp[i][i] = 1;
        }

        for (int i = s.length() - 1; i >= 0; i--) {
            for (int j = i + 1; j < s.length(); j++) {
                if (s.charAt(i) == s.charAt(j)) {
                    dp[i][j] = dp[i + 1][j - 1] + 2;
                } else {
                    dp[i][j] = Math.max(dp[i + 1][j], dp[i][j - 1]);
                }
            }
        }
        return dp[0][s.length() - 1];
    }
}

                

●  动态规划总结篇