LeetCode 39
1 class Solution { 2 public: 3 void dfs(int dep, int maxDep, vector<int>& cand, int target) 4 { 5 if (target < 0)return; 6 if (dep == maxDep) 7 { 8 if (target == 0)//到达尾部且等于target 9 { 10 vector<int> temp; 11 for (int i = 0; i < maxDep; i++) 12 { 13 for (int j = 0; j < num[i]; j++) 14 temp.push_back(cand[i]); 15 } 16 ret.push_back(temp); 17 } 18 return; 19 } 20 for (int i = 0; i <= target / cand[dep]; i++)//枚举合适的情况 21 { 22 num[dep] = i; 23 dfs(dep + 1, maxDep, cand, target - i*cand[dep]); 24 } 25 } 26 vector<vector<int>> combinationSum(vector<int>& candidates, int target) 27 { 28 int n = candidates.size(); 29 if (n == 0)return ret; 30 sort(candidates.begin(), candidates.end()); 31 num.resize(n); 32 ret.clear(); 33 dfs(0, n, candidates, target); 34 return ret; 35 36 } 37 private: 38 vector<vector<int>> ret;//返回的结果 39 vector<int> num;//用来存储每个数字的次数 40 };
LeetCode 40
1 class Solution { 2 public: 3 void dfs(int dep, int maxDep, vector<int>& cand, int target) 4 { 5 if (target < 0)return; 6 if (dep == maxDep) 7 { 8 if (target == 0) 9 { 10 vector<int> temp; 11 for (int i = 0; i < maxDep; i++) 12 { 13 for (int j = 0; j < num[i]; j++) 14 temp.push_back(cand[i]); 15 } 16 res.insert(temp); 17 } 18 return; 19 } 20 for (int i = 0; i<=1; i++) 21 { 22 num[dep] = i; 23 dfs(dep + 1, maxDep, cand, target - i*cand[dep]); 24 } 25 } 26 vector<vector<int>> combinationSum2(vector<int>& candidates, int target) 27 { 28 int n = candidates.size(); 29 if (n == 0)return ret; 30 sort(candidates.begin(), candidates.end()); 31 num.resize(n); 32 ret.clear(); 33 dfs(0, n, candidates, target); 34 set<vector<int>>::iterator it= res.begin(); 35 for (;it!=res.end();it++) 36 { 37 ret.push_back(*it); 38 } 39 return ret; 40 41 } 42 43 private: 44 vector<vector<int>> ret; 45 set<vector<int>> res; 46 vector<int> num; 47 };
LeetCode 22
backtracking函数书写的一般规则：
(1)函数参数一般要包括位置或者其它(如本题中的还可以剩余左括号个数及左边有多少个左括号没有关闭),这些都是为函数内容作为判断条件，要选择好。
(2)函数开头是函数终止条件(如本题中已经没有左括号可以使用了，故return）,并将结束得到的一个结果（元素可以不取的话，则将临时结果变量作为函数的参数<如subset>，每个元素都要取一个结果，则作为类成员）放入result中
(3)就是函数的后半部分，为递归，函数该位置可以放那些元素(循环)，每种元素放入后递归下一个位置的元素，当然包括参数的变化。
(1)参数：pos:迭代位置，共有2*n个元素，表示迭代到第几个位置了, left表示左边已经有多少个左括号没有关闭，当left==0时就不能迭代右括号了，remain表示还可以加入左括号的个数，当remain==0的时候，就不能迭代左括号了。
(2)终止条件：当remain==0时，只能加入右括号了，并将这一结果加入最后的结果中
(3)每个位置都只能放入左右括号两种情况，分别取这两种情况，然后进行下一次迭代。
1 class Solution { 2 public: 3 void dfs(int pos, string &str, int left, int remain, int n){ // left表示左边有多少个左括号没有关闭 4 if(remain == 0){ // 左括号使用完毕 故全部为右括号 5 for(int i = pos; i < 2*n; i++) 6 str[i] = ')'; 7 result.push_back(str); 8 return; 9 } 10 11 if(remain > 0){ // remain!=0 说明还可以使用左括号 12 str[pos] = '('; 13 dfs(pos+1, str, left+1, remain-1, n); 14 } 15 if(left != 0){ // left！=0 表示左边还有左括号没有关闭，故可以使用) 16 str[pos] = ')'; 17 dfs(pos+1, str, left-1, remain, n); 18 } 19 } 20 vector<string> generateParenthesis(int n) { 21 string str; 22 str.resize(2*n); 23 dfs(0, str, 0, n, n); 24 return result; 25 26 } 27 private: 28 vector<string> result; 29 };
Leetcode 77
1 class Solution { 2 private: 3 vector<vector<int>> result; 4 vector<int> tmp; 5 public: 6 void dfs(int dep, const int &n, int k, int count) 7 { 8 if (count == k) 9 { 10 result.push_back(tmp); 11 return; 12 } 13 if (dep < n - k + count) 14 { 15 dfs(dep + 1, n, k, count);//不取当前数 16 } 17 tmp[count] = dep + 1; //取当前数 18 dfs(dep + 1, n, k, count+1); 19 20 } 21 vector<vector<int>> combine(int n, int k) 22 { 23 if (k>n)k = n; 24 tmp.resize(k); 25 dfs(0, n, k, 0); 26 return result; 27 } 28 29 };
LeetCode 17
1 string NumToStr[10] = {"", "","abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"}; 2 class Solution { 3 private: 4 string str; 5 vector<string> result; 6 public: 7 int charToInt(char c) 8 { 9 int i; 10 stringstream ss; 11 ss << c; 12 ss >> i; 13 return i; 14 } 15 void dfs(int dep, int maxDep, const string &digits) 16 { 17 if (dep == maxDep) 18 { 19 result.push_back(str); 20 } 21 int index = charToInt(digits[dep]); 22 for (int i = 0; i < NumToStr[index].size(); i++) 23 { 24 str[dep] = NumToStr[index][i]; 25 dfs(dep+1,maxDep,digits); 26 } 27 } 28 29 vector<string> letterCombinations(string digits) 30 { 31 int n = digits.size(); 32 if (n == 0)return result; 33 str.resize(n); 34 dfs(0,n,digits); 35 return result; 36 } 37 };
LeetCode 401
1 class Solution { 2 vector<int> hour = { 1, 2, 4, 8 }, minute = {1,2,4,8,16,32}; 3 public: 4 void helper(vector<string> & res, pair<int, int> time, int num, int start_point) 5 { 6 if (num == 0) 7 { 8 if (time.second < 10) 9 res.push_back(to_string(time.first) + ":0" + to_string(time.second)); 10 else 11 res.push_back(to_string(time.first) + ":" + to_string(time.second)); 12 return; 13 } 14 for (int i = start_point; i < hour.size() + minute.size(); i++) 15 { 16 if (i < hour.size()) 17 { 18 time.first += hour[i]; 19 if (time.first < 12) 20 helper(res, time, num - 1, i + 1); 21 time.first -= hour[i]; 22 } 23 else 24 { 25 time.second += minute[i - hour.size()]; 26 if (time.second < 60) 27 helper(res,time,num-1,i+1); 28 time.second -= minute[i - hour.size()]; 29 } 30 } 31 } 32 vector<string> readBinaryWatch(int num) 33 { 34 vector<string> res; 35 helper(res,make_pair(0,0),num,0); 36 return res; 37 } 38 39 };
LeetCode 78
1 class Solution { 2 public: 3 vector<int>ans; 4 vector<vector<int>> res; 5 vector<vector<int>> subsets(vector<int>& nums) { 6 if(nums.empty())return res; 7 sort(nums.begin(),nums.end()); 8 dfs(0,ans,nums); 9 return res; 10 } 11 void dfs(int k,vector<int> ans, vector<int> nums) 12 { 13 res.push_back(ans); 14 for(int i=k;i<nums.size();i++) 15 { 16 ans.push_back(nums[i]); 17 dfs(i+1,ans,nums); 18 ans.pop_back(); 19 } 20 } 21 };
LeetCode 51
1 class Solution { 2 private: 3 vector<vector<string> > res; 4 public: 5 vector<vector<string> > solveNQueens(int n) { 6 vector<string>cur(n, string(n,'.')); 7 helper(cur, 0); 8 return res; 9 } 10 void helper(vector<string> &cur, int row) 11 { 12 if(row == cur.size()) 13 { 14 res.push_back(cur); 15 return; 16 } 17 for(int col = 0; col < cur.size(); col++) 18 if(isValid(cur, row, col)) 19 { 20 cur[row][col] = 'Q'; 21 helper(cur, row+1); 22 cur[row][col] = '.'; 23 } 24 } 25 26 //判断在cur[row][col]位置放一个皇后，是否是合法的状态 27 //已经保证了每行一个皇后，只需要判断列是否合法以及对角线是否合法。 28 bool isValid(vector<string> &cur, int row, int col) 29 { 30 //列 31 for(int i = 0; i < row; i++) 32 if(cur[i][col] == 'Q')return false; 33 //右对角线(只需要判断对角线上半部分，因为后面的行还没有开始放置) 34 for(int i = row-1, j=col-1; i >= 0 && j >= 0; i--,j--) 35 if(cur[i][j] == 'Q')return false; 36 //左对角线(只需要判断对角线上半部分，因为后面的行还没有开始放置) 37 for(int i = row-1, j=col+1; i >= 0 && j < cur.size(); i--,j++) 38 if(cur[i][j] == 'Q')return false; 39 return true; 40 } 41 };
LeetCode 46
1 class Solution { 2 private: 3 vector<vector<int>> result; 4 vector<int> visited; 5 public: 6 void dfs(int st, int n, vector<int>& v, vector<int>& nums) 7 { 8 if (st == n) 9 { 10 result.push_back(v); 11 return; 12 } 13 for (int i = 0; i < n; i++) 14 { 15 if (visited[i] != 1) 16 { 17 visited[i] = 1; 18 v.push_back(nums[i]); 19 dfs(st + 1, n,v,nums); 20 v.pop_back(); 21 visited[i] = 0; 22 } 23 } 24 } 25 vector<vector<int>> permute(vector<int>& nums) 26 { 27 int num = nums.size(); 28 visited.resize(num); 29 vector<int> tmp; 30 dfs(0,num,tmp,nums); 31 return result; 32 } 33 34 };
LeetCode 47
1 class Solution { 2 private: 3 vector<vector<int>> result; 4 vector<int> tmpResult; 5 vector<int> count; 6 public: 7 void dfs(int dep, int maxDep, vector<int>& num, vector<int> visited) 8 { 9 if (dep == maxDep) 10 { 11 result.push_back(tmpResult); 12 return; 13 } 14 for (int i = 0; i < num.size(); i++) 15 { 16 if (i == 0)count[dep] = 0; 17 if (!visited[i]) 18 { 19 count[dep]++; 20 if (count[dep]>1 && tmpResult[dep] == num[i])continue; 21 // 每个位置第二次选数时和第一次一样则continue 22 visited[i] = 1; // 每个位置第二次选择的数时和第一次不一样 23 tmpResult[dep] = num[i]; 24 dfs(dep + 1, maxDep, num, visited); 25 visited[i] = 0; 26 } 27 } 28 } 29 vector<vector<int>> permuteUnique(vector<int> &num) 30 { 31 sort(num.begin(), num.end()); 32 tmpResult.resize(num.size()); 33 count.resize(num.size()); 34 vector<int> visited; 35 visited.resize(num.size()); 36 dfs(0, num.size(), num, visited); 37 return result; 38 } 39 };
LeetCode 79
1 class Solution { 2 public: 3 bool dfs(int xi, int yi, string &word, int index, vector<vector<char> > &board, const int &m, const int &n, int **visited){ 4 visited[xi][yi] = 1; // 该结点已经访问过了 5 if(index + 1 < word.size()){ 6 if(xi-1 >= 0 && visited[xi-1][yi]==0 && board[xi-1][yi] == word[index+1]){ 7 if(dfs(xi-1, yi, word, index+1, board, m, n, visited))return true; //深度遍历 8 visited[xi-1][yi] = 0; // 这条路行不通 设为未访问 以不影响下面的遍历 9 } 10 if(xi+1 <m && visited[xi+1][yi]==0 && board[xi+1][yi] == word[index+1]){ 11 if(dfs(xi+1, yi, word, index+1, board, m, n, visited))return true; 12 visited[xi+1][yi] = 0; 13 } 14 if(yi-1 >= 0 && visited[xi][yi-1]==0 && board[xi][yi-1] == word[index+1]){ 15 if(dfs(xi, yi-1, word, index+1, board, m, n,visited)) return true; 16 visited[xi][yi-1] = 0; 17 } 18 if(yi+1 < n && visited[xi][yi+1]==0 && board[xi][yi+1] == word[index+1]){ 19 if(dfs(xi, yi+1, word, index+1, board, m, n,visited)) return true; 20 visited[xi][yi+1] = 0; 21 } 22 return false; 23 }else return true; 24 } 25 26 void initVisited(int ** visited, const int &m, const int &n){ 27 for(int i = 0; i < m; i++) 28 memset(visited[i], 0, sizeof(int)*n); 29 } 30 bool exist(vector<vector<char> > &board, string word) { 31 int m = board.size(); 32 int n = board[0].size(); 33 int **visited = new int*[m]; 34 for(int i = 0; i < m; i++) 35 visited[i] = new int[n]; 36 37 for(int i = 0; i < m; i++){ // 找到其中的i和j 38 for(int j = 0; j < n; j++){ 39 if(word[0] == board[i][j]){ 40 initVisited(visited, m, n); 41 if(dfs(i, j, word, 0, board, m, n,visited)) return true; 42 } 43 } 44 } 45 for(int i = 0; i < m; i++) 46 delete []visited[i]; 47 delete []visited; 48 return false; 49 } 50 };