647. Palindromic Substrings

Two Pointers

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class Solution {
    public int countSubstrings(String s) {
        int count = 0;
        for (int i = 0; i < s.length(); i++) {
            count += tryExpand(s, i, i);
            count += tryExpand(s, i, i + 1);
        }

        return count;
    }

    private int tryExpand(String s, int i, int j) {
        int count = 0;
        while (i >= 0 && j < s.length() && s.charAt(i) == s.charAt(j)) {
            i--;
            j++;
            count++;
        }

        return count;
    }
}

DP

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class Solution {
    public int countSubstrings(String s) {
        int n = s.length();
        int count = 0;

        // 定义 dp[i][j] 为字符串 s[i][j] 是否为回文字符串
        boolean[][] dp = new boolean[n][n];

        for (int i = n - 1; i >= 0; i--) {
            dp[i][i] = true;
            count++;
            
            for (int j = i + 1; j < n; j++) {
                if (s.charAt(i) == s.charAt(j)) {
                    // 注意不要忘记处理两个字符的特殊情况
                    if (j - i + 1 == 2) {
                        dp[i][j] = true;
                    } else {
                        dp[i][j] = dp[i + 1][j - 1];
                    }
                }

                if (dp[i][j]) {
                    count++;
                }
            }
        }

        return count;
    }
}

References

647. Palindromic Substrings
剑指 Offer II 020. 回文子字符串的个数