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  • 73th LeetCode Weekly Contest Domino and Tromino Tiling

    We have two types of tiles: a 2x1 domino shape, and an "L" tromino shape. These shapes may be rotated.

    XX  <- domino
    
    XX  <- "L" tromino
    X
    

    Given N, how many ways are there to tile a 2 x N board? Return your answer modulo 10^9 + 7.

    (In a tiling, every square must be covered by a tile. Two tilings are different if and only if there are two 4-directionally adjacent cells on the board such that exactly one of the tilings has both squares occupied by a tile.)

    Example:
    Input: 3
    Output: 5
    Explanation: 
    The five different ways are listed below, different letters indicates different tiles:
    XYZ XXZ XYY XXY XYY
    XYZ YYZ XZZ XYY XXY

    Note:

    • N  will be in range [1, 1000].

    一块区域有--,L(上下两种位置),问2*N有几种放置方式。

    首先

    x 这种我们叫0

    x

    x这种我们叫1

    xx

    xx这种我们叫2

    x

    xxx是怎么得到的呢。一种是x+横着和竖着  xx+竖着 xx+L形(上下两种)

    xxx                                      x                     xx          x

    于是...dp[i][0]=dp[i-2][0]+dp[i-1][0]+dp[i-1][1]+dp[i-1][2] i表示列数啦

    xxxx是怎么得到的呢。xx+L   xx+ --

    xxx             xx    xxx

    dp[i][1]=dp[i-2][0]+dp[i-1][2]

    xxx同理

    xxxx

     1 class Solution {
     2 public:
     3     const int mod = 1000000007;
     4     int dp[2000][3];
     5     int numTilings(int N) {
     6         dp[1][0]=1,dp[0][0]=1;
     7         for(int i=2;i<=N;i++){
     8             dp[i][0]=(dp[i-2][0]%mod+dp[i-1][0]%mod)%mod+(dp[i-1][1]%mod+dp[i-1][2]%mod)%mod;
     9             dp[i][1]=(dp[i-2][0]%mod+dp[i-1][2]%mod)%mod;
    10             dp[i][2]=(dp[i-2][0]%mod+dp[i-1][1]%mod)%mod;
    11         }
    12         return dp[N][0]%mod;
    13     }
    14 };
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  • 原文地址:https://www.cnblogs.com/yinghualuowu/p/8503323.html
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