zoukankan      html  css  js  c++  java
  • LeetCode 576. Out of Boundary Paths

    原题链接在这里:https://leetcode.com/problems/out-of-boundary-paths/description/

    题目:

    There is an m by n grid with a ball. Given the start coordinate (i,j) of the ball, you can move the ball to adjacent cell or cross the grid boundary in four directions (up, down, left, right). However, you can at most move N times. Find out the number of paths to move the ball out of grid boundary. The answer may be very large, return it after mod 109 + 7.

    Example 1:

    Input:m = 2, n = 2, N = 2, i = 0, j = 0
    Output: 6
    Explanation:
    

    Example 2:

    Input:m = 1, n = 3, N = 3, i = 0, j = 1
    Output: 12
    Explanation:
    

    Note:

    1. Once you move the ball out of boundary, you cannot move it back.
    2. The length and height of the grid is in range [1,50].
    3. N is in range [0,50].

    题解:

    DP问题.

    初始化dp[x][y]表示到x,y这个位置上有多少种走法. 刚开始给的 i, j 位置上是1, 其他位置全是0. 

    状态转移 (x,y) 从上下左右四个方向 往外走,走到dx, dy. dp[x][y] 就加到dp[dx][dy]上.

    如果(dx, dy)已经出界了, 那说明这就是提出boundary了. 可以把dp[x][y] 加到res中.

    答案是在每一种状态时边界点能从不同方向走出来的和.

    Note: 角上只有一个点, 但可以从不同方向走出来,每个方向都要算一次.

    Time Complexity: O(N*m*n).

    Space: O(m*n).

    AC Java:

     1 class Solution {
     2     public int findPaths(int m, int n, int N, int i, int j) {
     3         int mod = 1000000007;
     4         int [][] dp = new int[m][n];
     5         dp[i][j] = 1;
     6         int res = 0;
     7         int [][] dirs = {{-1,0},{1,0},{0,-1},{0,1}};
     8         
     9         for(int move = 0; move<N; move++){
    10             int [][] temp = new int[m][n];
    11             for(int x = 0; x<m; x++){
    12                 for(int y = 0; y<n; y++){
    13                     for(int [] dir : dirs){
    14                         int dx = x+dir[0];
    15                         int dy = y+dir[1];
    16                         if(dx<0 || dx==m || dy<0 || dy==n){
    17                             res = (res+dp[x][y])%mod;
    18                         }else{
    19                             temp[dx][dy] = (temp[dx][dy]+dp[x][y])%mod;
    20                         }
    21                     }
    22                 }
    23             }
    24             dp = temp;
    25         }
    26         return res;
    27     }
    28 }

    类似Knight Probability in Chessboard.

  • 相关阅读:
    Fragment中获取Activity的Context (转)
    raw cannot be resolved or is not a field解决办法
    商业分析07_06分布情况分析
    商业分析07_04漏斗分析
    商业分析07_03数据涨跌异动如何处理
    商业分析07_02多维度拆解
    商业分析07_01对比分析
    商业分析07_00概述 数据分析
    商业分析06选择数据工具
    商业分析05如何选取数据指标
  • 原文地址:https://www.cnblogs.com/Dylan-Java-NYC/p/7595100.html
Copyright © 2011-2022 走看看