原题链接在这里:https://leetcode.com/problems/dice-roll-simulation/
题目:
A die simulator generates a random number from 1 to 6 for each roll. You introduced a constraint to the generator such that it cannot roll the number i
more than rollMax[i]
(1-indexed) consecutive times.
Given an array of integers rollMax
and an integer n
, return the number of distinct sequences that can be obtained with exact n
rolls.
Two sequences are considered different if at least one element differs from each other. Since the answer may be too large, return it modulo 10^9 + 7
.
Example 1:
Input: n = 2, rollMax = [1,1,2,2,2,3] Output: 34 Explanation: There will be 2 rolls of die, if there are no constraints on the die, there are 6 * 6 = 36 possible combinations. In this case, looking at rollMax array, the numbers 1 and 2 appear at most once consecutively, therefore sequences (1,1) and (2,2) cannot occur, so the final answer is 36-2 = 34.
Example 2:
Input: n = 2, rollMax = [1,1,1,1,1,1] Output: 30
Example 3:
Input: n = 3, rollMax = [1,1,1,2,2,3] Output: 181
Constraints:
1 <= n <= 5000
rollMax.length == 6
1 <= rollMax[i] <= 15
题解:
Let dp[i][j][k] denotes number of distinct sequences up to i dice, ending with number j with k repeating j at the end.
For previous p, if j != p, then dp[i][j][1] = sum(dp[i - 1][p][k]), sum of previous ending with p, with k repeating p at the end.
Else if j == p, we need to make sure k + 1 <= rollMax[j], dp[i][j][k + 1] = dp[i - 1][j][k]. otherwise it would be 0.
Time Complexity: O(n*m*m*rMax). m = 6. rMax = max(rollMax[i]).
Space: O(n*m*rMax).
AC Java:
1 class Solution { 2 public int dieSimulator(int n, int[] rollMax) { 3 int mod = 1000000007; 4 int rMax = 15; 5 6 int [][][] dp = new int[n + 1][6][rMax + 1]; 7 for(int i = 0; i < 6; i++){ 8 dp[1][i][1] = 1; 9 } 10 11 for(int i = 2; i <= n; i++){ 12 for(int j = 0; j < 6; j++){ 13 for(int p = 0; p < 6; p++){ 14 for(int k = 1; k <= rMax; k++){ 15 if(j != p){ 16 dp[i][j][1] = (dp[i][j][1] + dp[i - 1][p][k]) % mod; 17 }else if(k < rollMax[j]){ // same number, make sure k + 1 <= rollMax[j] 18 dp[i][j][k + 1] = dp[i - 1][j][k]; 19 } 20 } 21 } 22 } 23 } 24 25 int res = 0; 26 for(int j = 0; j < 6; j++){ 27 for(int k = 1; k <= rMax; k++){ 28 res = (res + dp[n][j][k]) % mod; 29 } 30 } 31 32 return res; 33 } 34 }
原文地址:https://www.cnblogs.com/Dylan-Java-NYC/p/12325988.html