代码随想录day44：动态规划over，回文子串及字序列

day44：动态规划over，回文子串

647.回文子串

class Solution {
    public int countSubstrings(String s) {
// 布尔类型的dp[i][j]：表示区间范围[i,j] （注意是左闭右闭）的子串是否是回文子串，
// 如果是dp[i][j]为true，否则为false
        char[] chars = s.toCharArray();
        int n = s.length();
        boolean[][] dp = new boolean[n][n];
        int ans = 0;
        for (int i = n - 1; i >= 0; i--) {
            for (int j = i; j < n; j++) {
                if (chars[i] == chars[j]) {
                    if (j - i <= 1 || dp[i + 1][j - 1]) {
                        ans++;
                        dp[i][j] = true;
                    }
                }
            }
        }
        return ans;
    }
}

516.最长回文子序列

class Solution {
    public int longestPalindromeSubseq(String s) {
// dp[i][j]：字符串s在[i, j]范围内最长的回文子序列的长度为dp[i][j]
        int n = s.length();
        char[] chars = s.toCharArray();
        int[][] dp = new int[n][n];
        for (int i = 0; i < n; i++) dp[i][i] = 1;
        for (int i = n - 1; i >= 0; i--) {
            for (int j = i + 1; j < n; j++) {
                if (chars[i] == chars[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][n - 1];
    }
}