1汉诺塔 

从左移到最右，圆盘必须满足小压大原则

写一个大方法，大方法包括两步：第一步将最后一个圆盘上面的所有的放到第二个塔上面，然后将最后一个圆盘放到最后塔上面，再把第二个塔上面圆盘全放在第三个塔上面

#include <iostream>using namespace std;// 将 1~N 层圆盘从左 -> 右
void leftToRight(int n);// 将 1~N 层圆盘从左 -> 中
void leftToMid(int n);// 将 1~N 层圆盘从右 -> 中
void rightToMid(int n);// 将 1~N 层圆盘从中 -> 右
void midToRight(int n);// 将 1~N 层圆盘从中 -> 左
void midToLeft(int n);// 将 1~N 层圆盘从右 -> 左
void rightToLeft(int n);// 汉诺塔问题
void hanoi1(int n) {leftToRight(n);
}void leftToRight(int n) {if (n == 1) { // 基本情况cout << "Move 1 from left to right" << endl;return;}leftToMid(n - 1);cout << "Move " << n << " from left to right" << endl;midToRight(n - 1);
}void leftToMid(int n) {if (n == 1) {cout << "Move 1 from left to mid" << endl;return;}leftToRight(n - 1);cout << "Move " << n << " from left to mid" << endl;rightToMid(n - 1);
}void rightToMid(int n) {if (n == 1) {cout << "Move 1 from right to mid" << endl;return;}rightToLeft(n - 1);cout << "Move " << n << " from right to mid" << endl;leftToMid(n - 1);
}void midToRight(int n) {if (n == 1) {cout << "Move 1 from mid to right" << endl;return;}midToLeft(n - 1);cout << "Move " << n << " from mid to right" << endl;leftToRight(n - 1);
}void midToLeft(int n) {if (n == 1) {cout << "Move 1 from mid to left" << endl;return;}midToRight(n - 1);cout << "Move " << n << " from mid to left" << endl;rightToLeft(n - 1);
}void rightToLeft(int n) {if (n == 1) {cout << "Move 1 from right to left" << endl;return;}rightToMid(n - 1);cout << "Move " << n << " from right to left" << endl;midToLeft(n - 1);
}int main() {// 测试int n = 3; // 圆盘的层数hanoi1(n);return 0;
}

 

递归2：把三个塔分为from other to 

#include <iostream>using namespace std;// 汉诺塔问题
void hanoi2(int n) {if (n > 0) {func(n, "left", "right", "mid");}
}// 递归函数
void func(int N, string from, string to, string other) {if (N == 1) { // 基本情况cout << "Move 1 from " << from << " to " << to << endl;} else {func(N - 1, from, other, to);cout << "Move " << N << " from " << from << " to " << to << endl;func(N - 1, other, to, from);}
}int main() {// 测试int n = 3; // 圆盘的层数hanoi2(n);return 0;
}

2打印一个字符串的全部子序列
递归思路：第一个参数是原字符串，第二个index是当前来到的字符，path是之前已经选好的部分串，第三个参数是所有结果的保存

#include <vector>
#include <string>using namespace std;class Solution {
public:vector<string> subs(string s) {vector<char> str(s.begin(), s.end());string path = "";vector<string> ans;process1(str, 0, ans, path);return ans;}private:void process1(vector<char>& str, int index, vector<string>& ans, string path) {if (index == str.size()) {ans.push_back(path);return;}// 不要索引位置的字符process1(str, index + 1, ans, path);// 要索引位置的字符process1(str, index + 1, ans, path + str[index]);}
};

 3题 求所有不同的子序列，只要改成set结构收集答案即可

4 打印一个字符串的全部排列

去除该字符后需要回溯去恢复现场

#include <vector>
#include <string>using namespace std;class Solution {
public:vector<string> permutation1(string s) {vector<string> ans;if (s.empty()) {return ans;}string path = "";vector<char> rest(s.begin(), s.end());f(rest, path, ans);return ans;}private:void f(vector<char>& rest, string path, vector<string>& ans) {if (rest.empty()) {ans.push_back(path);} else {int N = rest.size();for (int i = 0; i < N; i++) {char cur = rest[i];rest.erase(rest.begin() + i);f(rest, path + cur, ans);rest.insert(rest.begin() + i, cur);}}}
};

 递归2：

不停地做字符串前一位和后一位交换，而且是在原字符串上面交换

#include <vector>
#include <string>using namespace std;class Solution {
public:vector<string> permutation2(string s) {vector<string> ans;if (s.empty()) {return ans;}char* str = &s[0];g1(str, 0, ans);return ans;}private:void g1(char* str, int index, vector<string>& ans) {if (str[index] == '\0') {ans.push_back(string(str));} else {for (int i = index; str[i] != '\0'; i++) {swap(str[index], str[i]);g1(str, index + 1, ans);swap(str[index], str[i]);}}}void swap(char& a, char& b) {char temp = a;a = b;b = temp;}
};

5去重过后的排列

在上一题代码做改动，如果某个字符已经试过了，那就在后面在遇到时就不尝试了

 

#include <vector>
#include <string>using namespace std;class Solution {
public:vector<string> permutation3(string s) {vector<string> ans;if (s.empty()) {return ans;}char* str = &s[0];g2(str, 0, ans);return ans;}private:void g2(char* str, int index, vector<string>& ans) {if (str[index] == '\0') {ans.push_back(string(str));} else {bool visited[256] = { false }; // 记录字符是否已被访问for (int i = index; str[i] != '\0'; i++) {if (!visited[str[i]]) { // 如果字符尚未访问过visited[str[i]] = true; // 标记字符已被访问swap(str[index], str[i]);g2(str, index + 1, ans);swap(str[index], str[i]);}}}}void swap(char& a, char& b) {char temp = a;a = b;b = temp;}
};

 6 递归逆序栈

#include<stack>
using namespace std;
//该函数的功能是返回栈底的元素
int f(stack<int>& s)
{int result = s.top();s.pop();if (s.empty()){return result;}else {int last = f(s);s.push(result);return last;}
}
void reverse(stack<int>& s)
{if (s.empty())return;int i = f(s);reverse(s);s.push(i);
}