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

dpij 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 dpi - 1j - 1

if (si ==sj)

if （j - i <= 1) dpij = true; res++;

else if (dpi - 1j - 1 == true) dpij = true; res++;

初始化：dpij

dpij = 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

dpij i,j的回文子串并列长度为

if (si == sj) dpij = dpi + 1j - 1 + 2//左右都加了一个元素

else dpij = max(dpij - 1, dpi + 1j)

初始化： dpii = 1;

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

dpii = 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