README_EN.md

08.06. Hanota

中文文档

Description

In the classic problem of the Towers of Hanoi, you have 3 towers and N disks of different sizes which can slide onto any tower. The puzzle starts with disks sorted in ascending order of size from top to bottom (i.e., each disk sits on top of an even larger one). You have the following constraints:

(1) Only one disk can be moved at a time.
(2) A disk is slid off the top of one tower onto another tower.
(3) A disk cannot be placed on top of a smaller disk.

Write a program to move the disks from the first tower to the last using stacks.

Example1:

 Input: A = [2, 1, 0], B = [], C = []
 Output: C = [2, 1, 0]

Example2:

 Input: A = [1, 0], B = [], C = []
 Output: C = [1, 0]

Note:

1. A.length <= 14

Solutions

Python3

class Solution:
    def hanota(self, A: List[int], B: List[int], C: List[int]) -> None:
        def dfs(n, a, b, c):
            if n == 1:
                c.append(a.pop())
                return
            dfs(n - 1, a, c, b)
            c.append(a.pop())
            dfs(n - 1, b, a, c)

        dfs(len(A), A, B, C)

class Solution:
    def hanota(self, A: List[int], B: List[int], C: List[int]) -> None:
        stk = [(len(A), A, B, C)]
        while stk:
            n, a, b, c = stk.pop()
            if n == 1:
                c.append(a.pop())
            else:
                stk.append((n - 1, b, a, c))
                stk.append((1, a, b, c))
                stk.append((n - 1, a, c, b))

Java

class Solution {
    public void hanota(List<Integer> A, List<Integer> B, List<Integer> C) {
        dfs(A.size(), A, B, C);
    }

    private void dfs(int n, List<Integer> a, List<Integer> b, List<Integer> c) {
        if (n == 1) {
            c.add(a.remove(a.size() - 1));
            return;
        }
        dfs(n - 1, a, c, b);
        c.add(a.remove(a.size() - 1));
        dfs(n - 1, b, a, c);
    }
}

class Solution {
    public void hanota(List<Integer> A, List<Integer> B, List<Integer> C) {
        Deque<Task> stk = new ArrayDeque<>();
        stk.push(new Task(A.size(), A, B, C));
        while (stk.size() > 0) {
            Task task = stk.pop();
            int n = task.n;
            List<Integer> a = task.a;
            List<Integer> b = task.b;
            List<Integer> c = task.c;
            if (n == 1) {
                c.add(a.remove(a.size() - 1));
            } else {
                stk.push(new Task(n - 1, b, a, c));
                stk.push(new Task(1, a, b, c));
                stk.push(new Task(n - 1, a, c, b));
            }
        }
    }
}

class Task {
    int n;
    List<Integer> a;
    List<Integer> b;
    List<Integer> c;

    public Task(int n, List<Integer> a, List<Integer> b, List<Integer> c) {
        this.n = n;
        this.a = a;
        this.b = b;
        this.c = c;
    }
}

C++

class Solution {
public:
    void hanota(vector<int>& A, vector<int>& B, vector<int>& C) {
        function<void(int, vector<int>&, vector<int>&, vector<int>&)> dfs = [&](int n, vector<int>& a, vector<int>& b, vector<int>& c) {
            if (n == 1) {
                c.push_back(a.back());
                a.pop_back();
                return;
            }
            dfs(n - 1, a, c, b);
            c.push_back(a.back());
            a.pop_back();
            dfs(n - 1, b, a, c);
        };
        dfs(A.size(), A, B, C);
    }
};

struct Task {
    int n;
    vector<int>* a;
    vector<int>* b;
    vector<int>* c;
};

class Solution {
public:
    void hanota(vector<int>& A, vector<int>& B, vector<int>& C) {
        stack<Task> stk;
        stk.push({(int) A.size(), &A, &B, &C});
        while (!stk.empty()) {
            Task task = stk.top();
            stk.pop();
            if (task.n == 1) {
                task.c->push_back(task.a->back());
                task.a->pop_back();
            } else {
                stk.push({task.n - 1, task.b, task.a, task.c});
                stk.push({1, task.a, task.b, task.c});
                stk.push({task.n - 1, task.a, task.c, task.b});
            }
        }
    }
};

Go

func hanota(A []int, B []int, C []int) []int {
	var dfs func(n int, a, b, c *[]int)
	dfs = func(n int, a, b, c *[]int) {
		if n == 1 {
			*c = append(*c, (*a)[len(*a)-1])
			*a = (*a)[:len(*a)-1]
			return
		}
		dfs(n-1, a, c, b)
		*c = append(*c, (*a)[len(*a)-1])
		*a = (*a)[:len(*a)-1]
		dfs(n-1, b, a, c)
	}
	dfs(len(A), &A, &B, &C)
	return C
}

func hanota(A []int, B []int, C []int) []int {
	stk := []Task{{len(A), &A, &B, &C}}
	for len(stk) > 0 {
		task := stk[len(stk)-1]
		stk = stk[:len(stk)-1]
		if task.n == 1 {
			*task.c = append(*task.c, (*task.a)[len(*task.a)-1])
			*task.a = (*task.a)[:len(*task.a)-1]
		} else {
			stk = append(stk, Task{task.n - 1, task.b, task.a, task.c})
			stk = append(stk, Task{1, task.a, task.b, task.c})
			stk = append(stk, Task{task.n - 1, task.a, task.c, task.b})
		}
	}
	return C
}

type Task struct {
	n       int
	a, b, c *[]int
}

TypeScript

/**
 Do not return anything, modify C in-place instead.
 */
function hanota(A: number[], B: number[], C: number[]): void {
    const dfs = (n: number, a: number[], b: number[], c: number[]) => {
        if (n === 1) {
            c.push(a.pop()!);
            return;
        }
        dfs(n - 1, a, c, b);
        c.push(a.pop()!);
        dfs(n - 1, b, a, c);
    };
    dfs(A.length, A, B, C);
}

/**
 Do not return anything, modify C in-place instead.
 */
function hanota(A: number[], B: number[], C: number[]): void {
    const stk: any[] = [[A.length, A, B, C]];
    while (stk.length) {
        const [n, a, b, c] = stk.pop()!;
        if (n === 1) {
            c.push(a.pop());
        } else {
            stk.push([n - 1, b, a, c]);
            stk.push([1, a, b, c]);
            stk.push([n - 1, a, c, b]);
        }
    }
}

...