樓梯問題：輸出所有可能的走法

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樓梯有n階台階，上樓可以一步上1階，也可以一步上2階，編一程式列出每一種走法。 

     例：3階台階的走法是 
     { 
     { 1, 1, 1, }, 
     { 1, 2, }, 
     { 2, 1, }, 
      }

 

因為每一步只有兩個選擇1/2階，那麼根據題意可建立一棵二叉樹，比如當n=3時：

          0        /         1      2     / \     /    1  2    1   /  1

這樣（深搜+回溯）一棵樹的所有到達葉子節點的路徑就是上樓梯的所有方法了

（遞迴法）C++代碼如下：

#include <iostream>
#include <list>

struct Node{    int value = 0;    Node *left = nullptr;    Node *right = nullptr;};//@param n 總樓梯數//@param value 步長(1 || 2)//@param cur 當前已走過的總樓梯數void BuildTree(Node *root, int n, int value, int cur = 0){    root->value = value;    cur += value;    if (cur == n) return;    if (cur + 1 <= n)    {        root->left = new Node;        BuildTree(root->left, n, 1, cur);    }    if (cur + 2 <= n)    {        root->right = new Node;        BuildTree(root->right, n, 2, cur);    }}void PrintAll(Node *root, std::list<int> &vec, int cur = 0){    if (root == nullptr) return;    cur += root->value;    vec.push_back(root->value);    if (root->left == nullptr && root->right == nullptr)    {        for (int i : vec)        {            if(i != 0)                printf("%d ", i);        }        printf("\n");    }    PrintAll(root->left, vec, cur);    PrintAll(root->right, vec, cur);    vec.pop_back();}void DeleteAll(Node *root){    if (root == nullptr) return;    DeleteAll(root->left);    DeleteAll(root->right);    delete root;}void Compute(int n){    Node *root = new Node;    std::list<int> st;    BuildTree(root, n, 0);    PrintAll(root, st);    DeleteAll(root);}int main(){    std::cout << "n = 3 :" << std::endl;    Compute(3);    std::cout << std::endl;    std::cout << "n = 4:" << std::endl;    Compute(4);}

 

（遞迴法）lua 代碼：

--@param n 總樓梯數[email protected] value 步長(1 || 2)--@param cur 當前已走過的總樓梯數function BuildTree(root, n, value, cur)    root.value = value    cur = cur + value    if cur == n then return end    if cur + 1 <= n then        root.left = { value = 0, left = nil, right = nil}        BuildTree(root.left, n, 1, cur)    end    if cur + 2 <= n then        root.right = { value = 0, left = nil, right = nil}        BuildTree(root.right, n, 2, cur)    endend

function PrintAllPath(root, stack, cur)    if root == nil then return end    cur = cur + root.value    table.insert(stack, root.value)    if root.left == nil and root.right == nil then        for key, value in pairs(stack) do            if value ~= 0 then                io.write(value,‘ ‘)            end        end        io.write(‘\n‘)    end    PrintAllPath(root.left, stack, cur)    PrintAllPath(root.right, stack, cur)    table.remove(stack)endfunction Compute(n)    stack = {}    root = { value = 0, left = nil, right = nil}    BuildTree(root, n, 0, 0)    --BuildTree(root, n)    PrintAllPath(root, stack, 0)endprint("n=3:")Compute(3)io.write(‘\n‘)print("n=4:")Compute(4)io.write(‘\n‘)print("n=5:")Compute(5)

 

 

迭代法(C++):

void BuildTree(Node *root, int n){    if (root == nullptr) return;    int cur(0);    std::stack<Node*> st;    std::map<Node*, int> stepMap;//記錄當前已走過的樓梯數    Node *p = root;    st.push(p);    stepMap[p] = 0;    while (!st.empty())    {        p = st.top();        st.pop();        cur = stepMap[p];        if (cur + 2 <= n)        {            p->right = new Node;            p->right->value = 2;            if (cur + 2 < n)            {                st.push(p->right);                stepMap[p->right] = cur + 2;            }        }        if (cur + 1 <= n)        {            p->left = new Node;            p->left->value = 1;            if (cur + 1 < n)            {                st.push(p->left);                stepMap[p->left] = cur + 1;            }        }    }}

 

迭代法（Lua）：

function BuildTree(root, n)    if root == nil then return end    cur = 0    stack = {}    stepMap = {}    p = root    table.insert(stack, p)    stepMap[p] = 0    while(#stack ~= 0) do        p = stack[1]        table.remove(stack, 1)        cur = stepMap[p]        if cur + 2 <= n then            p.right = {value = 2, left = nil, right = nil}            if cur + 2 < n then                table.insert(stack, p.right)                stepMap[p.right] = cur + 2            end        end        if cur + 1 <= n then            p.left = {value = 1, left = nil, right = nil}            if cur + 1 < n then                table.insert(stack, p.left)                stepMap[p.left] = cur + 1            end        end    endend