代码随想录笔记---回溯篇

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回溯模板

void backtracking(参数)
{if(终止条件){存放结果;return;}for (选择:本层集合中的元素（树中节点孩子的数量就是集合的大小）){处理节点;backtracking(路径，选择列表); // 递归回溯,撤销处理结果;}
}

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一、2025.4.26

1. 学习内容

77. 组合

vector<vector<int>> result;vector<int> path;void backtracking(int n, int k, int startIndex){if (path.size() == k){result.push_back(path);return;}for (int i = startIndex; i <= n; i++){path.push_back(i);backtracking(n, k, i + 1);path.pop_back();}}vector<vector<int>> combine(int n, int k) {backtracking(n, k, 1);return result;}

77.组合(优化)

vector<vector<int>> result;vector<int> path;void backtracking(int n, int k, int startIndex){if (path.size() == k){result.push_back(path);return;}//进行剪枝的关键在于:列表中的剩余元素(n - i + 1) >= path数组需要存储的剩余数量 (k - path.size()),//进而得到了i <= n - (k - path.size()) + 1;for (int i = startIndex; i <= n - (k - path.size()) + 1; i++){path.push_back(i);backtracking(n, k, i + 1);path.pop_back();}}vector<vector<int>> combine(int n, int k) {backtracking(n, k, 1);return result;}

2.复习内容

707. 设计链表

class MyLinkedList {public:struct LinkedNode{int val;LinkedNode* next;LinkedNode(int x):val(x), next(nullptr){}};MyLinkedList() {dummy_Head = new LinkedNode(0);_size = 0;}int get(int index) {if (index > (_size - 1) || index < 0) return -1;LinkedNode* cur = dummy_Head -> next;while (index--){cur = cur -> next;}return cur -> val;}void addAtHead(int val) {LinkedNode* node = new LinkedNode(val);node -> next = dummy_Head -> next;dummy_Head -> next = node;_size++;}void addAtTail(int val) {LinkedNode* cur = dummy_Head;while (cur -> next){cur = cur -> next;}LinkedNode* node = new LinkedNode(val);cur -> next = node;_size++;}void addAtIndex(int index, int val) {if (index > _size) return;if (index < 0) index = 0;LinkedNode* cur = dummy_Head;while (index--){cur = cur -> next;}LinkedNode* node = new LinkedNode(val);node -> next = cur -> next;cur -> next = node;_size++;}void deleteAtIndex(int index) {if (index < -1 || index > (_size - 1)) return;LinkedNode* cur = dummy_Head;while (index--){cur = cur -> next;}LinkedNode* tmp = cur -> next;cur -> next = cur -> next -> next;delete tmp;_size--;}private:int _size;LinkedNode* dummy_Head;
};

206. 反转链表

ListNode* reverseList(ListNode* head) {ListNode* pre = nullptr, * cur = head, *tmp = nullptr;while (cur){tmp = cur -> next;cur -> next = pre;pre = cur;cur = tmp;}return pre;}

2025.5.1

1.学习内容

17. 电话号码的字母组合

const string letterMap[10] = {"","","abc","def","ghi","jkl","mno","pqrs","tuv","wxyz",};string path = "";vector<string> result;void backtracking(string digits, int index){if (index == digits.size()){result.push_back(path);return;}int digit = digits[index] - '0';for (int i = 0; i < letterMap[digit].size(); i++){string letter = letterMap[digit];path.push_back(letter[i]);backtracking(digits, index + 1);path.pop_back();}}vector<string> letterCombinations(string digits) {path.clear();result.clear();if (digits.size() == 0) return result;backtracking(digits, 0);return result;}

2.复习内容

216.组合总和III

vector<int> path;vector<vector<int>> result;void back_tracking(int startIndex, int sum, int n, int k){if (path.size() == k){if (sum == n) result.push_back(path);return;}if (sum >= n) return; //k - path.size() <= 9 - i + 1;for (int i = startIndex; i <= 9 - (k - path.size()) + 1; i++){path.push_back(i);sum += i;//记住回溯的条件选择:是i+1,而不是startIndex + 1back_tracking(i + 1, sum, n, k);sum -= i;path.pop_back();}}vector<vector<int>> combinationSum3(int k, int n) {path.clear();result.clear();back_tracking(1, 0, n, k);return result;}

2025.5.2

1.学习内容

39. 组合总和

vector<int> path;vector<vector<int>> result;void backtracking(vector<int> candidates, int target, int sum, int startIndex){if (sum > target) return;if (sum == target){result.push_back(path);return;}for (int i = startIndex; i < candidates.size(); i++){path.push_back(candidates[i]);sum += candidates[i];//相比于组合,这块应该是i + 1backtracking(candidates, target, sum, i);sum -= candidates[i];path.pop_back();}}vector<vector<int>> combinationSum(vector<int>& candidates, int target) {path.clear();result.clear();backtracking(candidates, target, 0, 0);return result;}

40. 组合总和 II

vector<int> path;vector<vector<int>> result;void backtracking(vector<int> candidates, int target, int sum, int startIndex, vector<bool> &used){if (sum > target) return;if (sum == target){result.push_back(path);return;}for (int i = startIndex; i < candidates.size(); i++){if (i > 0 && candidates[i] == candidates[i - 1] && used[i - 1] == false) continue;path.push_back(candidates[i]);sum += candidates[i];used[i] = true;backtracking(candidates, target, sum, i + 1, used);used[i] = false;sum -= candidates[i];path.pop_back();}}vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {vector<bool> used(candidates.size(), false);path.clear();result.clear();//感觉要将candidates数组进行排序sort(candidates.begin(), candidates.end());backtracking(candidates, target, 0, 0, used);return result;}

2025.5.7

1.学习内容

90. 子集 II

//子集问题就是要保存树的所有节点vector<int> path;vector<vector<int>> result;void backtracking(vector<int> nums, int startIndex, vector<bool>& used){//直接将节点存储到result中result.push_back(path);//省略终止条件,因为for循环结束后会自动终止for (int i = startIndex; i < nums.size(); i++){//used[i] == false应该就是树层去重if (i > 0 && nums[i - 1] == nums[i] && used[i - 1] == false) continue;path.push_back(nums[i]);used[i] = true;backtracking(nums, i + 1, used);used[i] = false;path.pop_back();}}vector<vector<int>> subsetsWithDup(vector<int>& nums) {path.clear();result.clear();vector<bool> used(nums.size(), false);sort(nums.begin(), nums.end());backtracking(nums, 0, used);return result;}

491.递增子序列

//使用哈希表
vector<int> path;vector<vector<int>> result;void backtracking(vector<int> nums, int startIndex){if (path.size() > 1) result.push_back(path);unordered_set<int> uset;for (int i = startIndex; i < nums.size(); i++){//排除情况: 1.不是递增; 2.在该层已经使用过了num[i]元素if ((path.size() > 0 && nums[i] < path.back()) || uset.find(nums[i]) != uset.end()) continue;uset.insert(nums[i]);path.push_back(nums[i]);backtracking(nums, i + 1);path.pop_back();}}vector<vector<int>> findSubsequences(vector<int>& nums) {path.clear();result.clear();backtracking(nums, 0);return result;}

使用数组代替字典
vector<int> path;vector<vector<int>> result;void backtracking(vector<int> nums, int startIndex){if (path.size() > 1) result.push_back(path);//由于-100 < nums[i] < 100,使用数组代替字典int uset[201] = {0};for (int i = startIndex; i < nums.size(); i++){//排除情况: 1.不是递增; 2.在该层已经使用过了num[i]元素if ((path.size() > 0 && nums[i] < path.back()) || uset[nums[i] + 100] == 1) continue;uset[nums[i] + 100] = 1;path.push_back(nums[i]);backtracking(nums, i + 1);path.pop_back();}}vector<vector<int>> findSubsequences(vector<int>& nums) {path.clear();result.clear();backtracking(nums, 0);return result;}

46. 全排列

vector<int> path;vector<vector<int>> result;void backtracking(vector<int> nums, vector<bool> &used){//终止条件if (path.size() == nums.size()){result.push_back(path);return;}for (int i = 0; i < nums.size(); i++){if (used[i] == true) continue;used[i] = true;path.push_back(nums[i]);backtracking(nums, used);path.pop_back();used[i] = false;}}vector<vector<int>> permute(vector<int>& nums) {path.clear();result.clear();vector<bool> used(nums.size(), false);backtracking(nums, used);return result;}

47. 全排列 II

vector<int> path;vector<vector<int>> result;void backtracking(vector<int> nums, vector<bool> &used){//终止条件if (path.size() == nums.size()){result.push_back(path);return;}for (int i = 0; i < nums.size(); i++){//一定要好好区分是树枝上去重还是树层上去重if (used[i] == true || (i > 0 && nums[i - 1] == nums[i] && used[i - 1] == false)) continue;used[i] = true;path.push_back(nums[i]);backtracking(nums, used);path.pop_back();used[i] = false;}}vector<vector<int>> permuteUnique(vector<int>& nums) {path.clear();result.clear();vector<bool> used(nums.size(), false);//去重一定要对元素进行排序sort(nums.begin(), nums.end());backtracking(nums, used);return result;}

2.复习内容

78. 子集

vector<int> path;vector<vector<int>> result;void backtracking(vector<int> nums, int startIndex){result.push_back(path);//这个终止条件可有可无if (startIndex >= nums.size()) return;for (int i = startIndex; i < nums.size(); i++){path.push_back(nums[i]);backtracking(nums, i + 1);path.pop_back();}}vector<vector<int>> subsets(vector<int>& nums) {path.clear();result.clear();backtracking(nums, 0);return result;}

93. 复原 IP 地址

vector<string> result;bool isValid(string s, int start, int end){if (start > end) return false;if (s[start] == '0' && start != end) return false;int num = 0;for (int i = start; i <= end; i++){if (s[i] < '0' && s[i] > '9') return false;num = num * 10 + (s[i] - '0');if (num > 255) return false;}return true;}void backtracking(string& s, int startIndex, int pointNum){if (pointNum == 3){if (isValid(s, startIndex, s.size() - 1)) result.push_back(s);return;}for (int i = startIndex; i < s.size(); i++){if (isValid(s, startIndex, i)){s.insert(s.begin() + i + 1, '.');pointNum++;backtracking(s, i + 2, pointNum);pointNum--;s.erase(s.begin() + i + 1);}}}vector<string> restoreIpAddresses(string s) {result.clear();backtracking(s, 0, 0);return result;}

24. 两两交换链表中的节点

ListNode* swapPairs(ListNode* head) {ListNode* dummu_Head = new ListNode(0);dummu_Head -> next = head;ListNode* cur = dummu_Head;while (cur -> next != nullptr && cur -> next -> next != nullptr){ListNode* tmp = cur -> next;ListNode* tmp1 = cur -> next -> next -> next;cur -> next = cur -> next -> next;cur -> next -> next = tmp;cur -> next -> next -> next = tmp1;cur = tmp;}head = dummu_Head -> next;return head;}

2025.5.11

1.学习内容

332. 重新安排行程

//Hierholzer 算法，我感觉和卡哥的思想很类似，但是该算法的思想有点难理解
unordered_map<string, priority_queue<string, vector<string>, std::greater<string>>> umap;vector<string> result;void dfs(string curr){while (umap.count(curr) && umap[curr].size() > 0){string tmp = umap[curr].top();umap[curr].pop();dfs(tmp);}result.push_back(curr);}vector<string> findItinerary(vector<vector<string>>& tickets) {for (auto &it : tickets){umap[it[0]].push(it[1]);}dfs("JFK");reverse(result.begin(), result.end());return result;}

51. N 皇后

vector<vector<string>> result;void dfs(int n, int row, vector<string>& chessboard, vector<bool> & col, vector<bool>& dp, vector<bool>& udp){if (row == n){result.push_back(chessboard);return;}for (int i = 0; i < n; i++){if (col[i] && dp[row + i] && udp[n - row + i]){col[i] = dp[row + i] = udp[n - row +　i] = false;chessboard[row][i] = 'Q';dfs(n, row + 1, chessboard, col, dp, udp);chessboard[row][i] = '.';col[i] = dp[row + i] = udp[n - row +　i] = true;}}}vector<vector<string>> solveNQueens(int n) {vector<string> chessboard(n, string(n, '.'));vector<bool> col(n, true), dp(2 * n, true), udp(2 * n, true);result.clear();dfs(n, 0, chessboard, col, dp, udp);return result;}

37. 解数独

//一定要使用实参（&），否则的话会创建大量副本，导致超时bool isValid(vector<vector<char>>& board, int row, int col, char c){//横向是否合适for (int i = 0; i < 9; i++){if (board[row][i] == c) return false;}//纵向是否合适for (int j = 0; j < 9; j++){if (board[j][col] == c) return false;}//九宫格内是否合适int startRow = (row / 3) * 3;int startCol = (col / 3) * 3;for (int i = startRow; i < startRow + 3; i++){for (int j = startCol; j < startCol + 3; j++){if (board[i][j] == c) return false;}}return true;}bool backtracking(vector<vector<char>>& board){//循环结束就终止，不需要终止条件for (int i = 0; i < board.size(); i++){for (int j = 0; j < board[0].size(); j++){if (board[i][j] == '.'){for (char c = '1'; c <= '9'; c++){if (isValid(board, i, j, c)){board[i][j] = c;if (backtracking(board)) return true;board[i][j] = '.';}}//9个数试完没有返回true，那就是falsereturn false;}}}return true;}void solveSudoku(vector<vector<char>>& board) {backtracking(board);}

70. 爬楼梯

//动态规划int climbStairs(int n) {vector<int> dp(n + 1, 0);if (n == 1) return 1;if (n == 2) return 2;dp[0] = 0;dp[1] = 1;dp[2] = 2;for (int i = 3; i <= n; i++) dp[i] = dp[i - 1] + dp[i - 2];return dp[n];}

//递归思路，但是当n大于40时会有明显的卡顿
int climbStairs(int n) {if (n <= 2) return n;return climbStairs(n - 1) + climbStairs(n - 2);}