解题思路见注释
/*** @author dustdawn* @date 2019/9/2 8:24*/
public class JiGuChuanHua {public static void main(String[] args) {int n = 3; //人数int m = 3; //次数transmit(n, m);}public static void transmit(int n, int m) {int[][] arr = new int[m+1][n+1];/*** 定义 行:次数+1 列:人数+1 的二维数组* 表示有n个人,第m次回到原点左右两人的情况之和* 停留在每个位置有两个方向走下一步* 即1->2 1->n* 则走一步回到一个点有两种情况2->1 n->1* arr[m][n]等于 m-1次走回原点之和* arr[m][n] = arr[m-1][n+1] + arr[m-1][n-1]* arr[0][1] = 1;* arr[1][2] = 1 = arr[0][1] + arr[0][3](0);* arr[2][2] = 1 = arr[1][1] + arr[1][3];* arr[m][2] = 1 = arr[m-1][1] + arr[m-1][3];*///走1次回到原点方式,只能1个人arr[0][1] = 1;//递归回边界arr[0][1]for (int i = 1; i <= m; i++) {for (int j = 1; j <= n; j++) {if(j == 1) {arr[i][1] = arr[i-1][2] + arr[i-1][n];}else if(j == n) {arr[i][j] = arr[i-1][j-1] + arr[i-1][1];}else {arr[i][j] = arr[i-1][j-1] + arr[i-1][j+1];}}}System.out.println(arr[m][1]);}}