576. Out of Boundary Paths

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:

Once you move the ball out of boundary, you cannot move it back.

The length and height of the grid is in range [1,50].

N is in range [0,50].

思路：
由于每次dp[step][i][j]只与dp[step-1][i][j]有关，所以可以优化空间复杂度，用一个二维数组temp记录上一步的状态。

class Solution
    public int findPaths(int m, int n, int N, int i, int j) {
        if (N <= 0) 
            return 0;
        final int MOD = 1000000007;
        int[][] count = new int[m][n];
        count[i][j] = 1;
        int result = 0;
        int[][] dirs = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
        for (int step = 0; step < N; step++) {
           int[][] temp = new int[m][n];
           for (int r = 0; r < m; r++) {
               for (int c = 0; c < n; c++) {
                   for (int[] d : dirs) {
                       int nr = r + d[0];
                       int nc = c + d[1];
                       if (nr < 0 || nr >= m || nc < 0 || nc >= n) {
                           result = (result + count[r][c]) % MOD;
                       } else {
                           temp[nr][nc] = (temp[nr][nc] + count[r][c]) % MOD;
                       }
                   }
               }

           }
           count = temp;
        }
        return