Java - 13 方法的递归

Java - 13 方法的递归

递归必须向退出递归的条件逼近

斐波那契数列

1. 当n=1，1
2. 当n=2，1
3. 当n >= 3，是前两个数的和

public class Recursion{public static void main(String[] args) {Fbnq f = new Fbnq();int n = 7;int res = f.num(n);if(res!=-1)System.out.println(res);}
}class Fbnq {public int num(int n){if(n<1){System.out.println("text the number >= 1");return -1;			}else{if(n > 2)return num(n-1)+num(n-2);elsereturn 1;}}
}

• 考虑无效输入的情况，返回-1，用-1控制输出

猴子吃桃

1. 第10天，1个桃子
2. 第9天，(1+1)*2个桃子

public class Recursion{public static void main(String[] args) {Monkey m = new Monkey();int res = m.peach(9);if (res!=-1)System.out.println(res);}
}class Monkey {public int peach(int n){if(n == 10){return 1;}else if(n>=1&&n<=9){return (peach(n+1)+1)*2;}else{System.out.println("1-10");return -1;}}
}

迷宫

public class Migong{public static void main(String[] args) {// draw mapint[][] map = new int[8][7];for(int i = 0; i<map[i].length; i++){map[0][i] = 1;map[7][i] = 1;}for(int i = 0; i<map.length; i++){map[i][0] = 1;map[i][6] = 1;}map[3][1] = 1;map[3][2] = 1;map[2][2] = 1;for(int i = 0; i<map.length; i++){for(int j = 0; j<map[i].length; j++){System.out.print(map[i][j]+" ");}System.out.println("");}// find wayMigong m = new Migong();m.findWay(map, 1, 1);System.out.println("============");// check the mapfor(int i = 0; i<map.length; i++){for(int j = 0; j<map[i].length; j++){System.out.print(map[i][j]+" ");}System.out.println("");}}
}class Migong {/* 1: barrier 2: can pass 3: cant pass */public boolean findWay(int[][] map, int i, int j){if(map[6][5] == 2)return true;else if(map[i][j] == 0){map[i][j] = 2; // assume it can go through// go downif(findWay(map, i+1, j))return true;// go rightelse if(findWay(map, i, j+1))return true;// go upelse if(findWay(map, i-1, j))return true;// go leftelse if(findWay(map, i, j-1))return true;else{map[i][j] = 3;return false;}}else{return false;}}
}

Hanoi

public class Hanoi{public static void main(String[] args) {Hanoi tower = new Hanoi();tower.move(5,'A','C','B');}
}class Hanoi {public void move(int num, char from, char to, char help){if(num == 1)System.out.println(from + "->" + to);else{move(num-1, from, help, to);System.out.println(from + "->" + to);move(num-1, help, to, from);}}
}

八皇后

import java.util.ArrayList;
import java.util.List;
public class Try{public static void main(String[] args) {int n = 8;Solution s = new Solution();s.solveNQueens(n);}
}
class Solution {public void solveNQueens(int n) {int count = 0;int[] solve = new int[n];for(int i = 0; i<n; i++){solve[i] = -1;}List<int[]> solves = new ArrayList<>();findQ(n, 0, solve, solves);// 遍历 Listfor (int[] array : solves) {count++;System.out.print("solution: ");for (int num : array) {System.out.print(num + " ");}System.out.println();}System.out.println(count); // 一共解法}public void findQ(int n, int row, int[] solve, List<int[]> solves){ // n:一共的行数(8); row:当前的行数(0-7); solvefor(int i = 0; i<n; i++){if(row == n){ // 最后一行已经下好Q了solves.add(solve.clone()); // *return ;}else{solve[row] = i; // 假设下在第row+1行的i+1个位置if(isValid(solve, row))findQ(n, row+1, solve, solves); // 继续找下一行solve[row] = -1; // 回溯}    }}public boolean isValid(int[] sol, int row){boolean flag1;boolean flag2;boolean flag3;for(int n = 0; n<row ; n++){for(int m = n+1; m<=row; m++){flag1 = sol[n] == sol[m]; // 在同一列flag2 = (n - sol[n]) == (m - sol[m]); // 在左上\右下flag3 = (n + sol[n]) == (m + sol[m]); // 在左下\右上if(flag1 || flag2 || flag3)return false;}}return true;}    }