LeetCode :647. Palindromic Substrings 回文子串数量-蒲公英云

LeetCode :647. Palindromic Substrings 回文子串数量

试题
Given a string, your task is to count how many palindromic substrings in this string.

The substrings with different start indexes or end indexes are counted as different substrings even they consist of same characters.

Example 1:

Input: “abc”
Output: 3
Explanation: Three palindromic strings: “a”, “b”, “c”.

Example 2:

Input: “aaa”
Output: 6
Explanation: Six palindromic strings: “a”, “a”, “a”, “aa”, “aa”, “aaa”.

Note:

The input string length won’t exceed 1000.

代码
想复杂了，其实这个按中点展开没什么区别。

class Solution {
    public int countSubstrings(String s) {
        int len = s.length();
        char[] ss = s.toCharArray();
        int count = 0;
        int[][] dp = new int[len][len];
        for(int i=0; i<len; i++){
            dp[i][i] = 1;
            count += 1;
            if(i+1<len && ss[i]==ss[i+1]){
                dp[i][i+1] = 1;
                count ++;
            }
        }
        for(int g=2; g<len; g++){
            for(int i=0; i+g<len; i++){
                int sta=i, end=i+g;
                if(ss[sta]==ss[end] && dp[sta+1][end-1]==1){
                    dp[sta][end] = 1;
                    count += 1;
                }
            }
        }
        return count;
    }
}

中点展开代码：

class Solution {
    public int countSubstrings(String S) {
        int N = S.length(), ans = 0;
        for (int center = 0; center <= 2*N-1; ++center) {
            int left = center / 2;
            int right = left + center % 2;
            while (left >= 0 && right < N && S.charAt(left) == S.charAt(right)) {
                ans++;
                left--;
                right++;
            }
        }
        return ans;
    }
}

O(n)代码：

class Solution {
    public int countSubstrings(String S) {
        char[] A = new char[2 * S.length() + 3];
        A[0] = '@';
        A[1] = '#';
        A[A.length - 1] = '$';
        int t = 2;
        for (char c: S.toCharArray()) {
            A[t++] = c;
            A[t++] = '#';
        }
        int[] Z = new int[A.length];
        int center = 0, right = 0;
        for (int i = 1; i < Z.length - 1; ++i) {
            if (i < right)
                Z[i] = Math.min(right - i, Z[2 * center - i]);
            while (A[i + Z[i] + 1] == A[i - Z[i] - 1])
                Z[i]++;
            if (i + Z[i] > right) {
                center = i;
                right = i + Z[i];
            }
        }
        int ans = 0;
        for (int v: Z) ans += (v + 1) / 2;
        return ans;
    }
}