递归)"/>
骑士在棋盘上的概率(递归)
这道题使用递归思想虽然代码较简洁,但会出现大量的重复计算,但还是想把这种思路展示给大家:
class Solution {
public://vector<int> dirs = {{-2,1},{-1,2},{1,2},{2,1},{2,-1},{1,-2},{-1,-2},{-2,-1}};double knightProbability(int n, int k, int row, int column) {return (double)process(n, k, row, column)/pow(8,k);}long process(int n, int k, int row, int column){if(row<0||column<0||row>=n||column>=n){return 0;}if(k==0){return 1;}long d1 = process(n, k-1, row-2, column+1);long d2 = process(n, k-1, row-1, column+2);long d3 = process(n, k-1, row+1, column+2);long d4 = process(n, k-1, row+2, column+1);long d5 = process(n, k-1, row+2, column-1);long d6 = process(n, k-1, row+1, column-2);long d7 = process(n, k-1, row-1, column-2);long d8 = process(n, k-1, row-2, column-1);return d1 + d2 + d3 + d4 + d5 + d6 + d7 + d8;}
};
在n和k比较大的时候,栈会溢出,因此要转换思路使用动态规划求解,力扣上有官方题解,这里就不在重复了。
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骑士在棋盘上的概率(递归)
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