LWC 59：730. Count Different Palindromic Subsequences
LWC 59：730. Count Different Palindromic Subsequences
LWC 59：730. Count Different Palindromic Subsequences
传送门：730. Count Different Palindromic Subsequences
Problem:
Given a string S, find the number of different non-empty palindromic subsequences in S, and return that number modulo 10^9 + 7. A subsequence of a string S is obtained by deleting 0 or more characters from S. A sequence is palindromic if it is equal to the sequence reversed. Two sequences A_1, A_2, … and B_1, B_2, … are different if there is some i for which A_i != B_i.
Example 1:
Input: S = ‘bccb’ Output: 6 Explanation: The 6 different non-empty palindromic subsequences are ‘b’, ‘c’, ‘bb’, ‘cc’, ‘bcb’, ‘bccb’. Note that ‘bcb’ is counted only once, even though it occurs twice.
Example 2:
Input: S = ‘abcdabcdabcdabcdabcdabcdabcdabcddcbadcbadcbadcbadcbadcbadcbadcba’ Output: 104860361 Explanation: There are 3104860382 different non-empty palindromic subsequences, which is 104860361 modulo 10^9 + 7.
Note:
The length of S will be in the range [1, 1000].
Each character S[i] will be in the set {‘a’, ‘b’, ‘c’, ‘d’}.
思路： 难点在于如何划分子问题，才能保证更新dp时没有重复，其中需要解决重复元素子串的表达。为了保证每个子问题的回文在原问题中没有出现过，定义如下规则：子问题求出的回文串必须套上一层外壳，即子问题中的回文串集合Set = ｛s | s 为回文｝， 有新的回文 s’ = “a” + s + “a” or “b” + s + “b”,….
定义函数如下f(i, j) 表示当前对应S[i,…j]的不重复回文串个数，于是有：
初始化： ans = 0
1. 子问题的回文串批层外衣，有 ans += f(i + 1, j - 1) , 其中S[i] == S[j]
2. 考虑"a_..._a", "_..._"表示子问题的回文串，抽出a'= a...a，其中"..."表示x个a，那么有新的回文串aa...a 和 aa...aa,有ans += 2代码如下：
public int countPalindromicSubsequences(String S) {
int n = S.length();
int[][] next = new int[4][1010];
int[][] prev = new int[4][1010];
char[] cs = S.toCharArray();
for (int i = 0; i < 4; ++i) Arrays.fill(next[i], n);
for (int i = n - 1; i >= 0; --i) {
int c = cs[i] - 'a';
for (int j = 0; j < 4; ++j) next[j][i] = i + 1 == n ? n : next[j][i + 1];
next[c][i] = i;
}
for (int i = 0; i < 4; ++i) Arrays.fill(prev[i], -1);
for (int i = 0; i < n; ++i) {
int c = cs[i] - 'a';
for (int j = 0; j < 4; ++j) prev[j][i] = i - 1 == -1 ? -1 : prev[j][i - 1];
prev[c][i] = i;
}
dp = new int[1010][1010];
return f(cs, next, prev, 0, n - 1);
}
int mod = 1000000000 + 7;
int[][] dp;
int f(char[] cs, int[][] next, int[][] prev, int s, int e) {
if (s > e) return 0;
if (dp[s][e] > 0) return dp[s][e];
long ans = 0;
for (int i = 0; i < 4; ++i) {
int ns = next[i][s];
int ne = prev[i][e];
if (ns > ne) continue;
if (ns != ne) ans += 1;
ans ++;
ans += f(cs, next, prev, ns + 1, ne - 1);
}
dp[s][e] = (int)(ans % mod);
return dp[s][e];
} 腾讯云开发者
