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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falsechar[] 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;elsedp[i][j] = Math.max(dp[i + 1][j], dp[i][j - 1]);}}return dp[0][n - 1];}
}