因为 一刷 也不求大家能把这么难的问题解决，所以 大家一刷的时候，就了解一下题目的要求，了解一下解题思路，不求能直接写出代码，先大概熟悉一下这些题，二刷的时候，随着对回溯算法的深入理解，再去解决如下三题。 

大家今天的任务，其实是 对回溯算法章节做一个总结就行。 

重点是看 回溯算法总结篇：

代码随想录

332.重新安排行程（可跳过） 

代码随想录

Python使用used数组，lc1/81会超时:

class Solution:def __init__(self):self.result = []self.path = ["JFK"]def backtracking(self, tickets, cur,  used):if len(self.path)==len(tickets)+1:self.result = self.path[:]return Truefor i, ticket in enumerate(tickets):if ticket[0] == cur and used[i] == False:used[i] = True    self.path.append(ticket[1])state = self.backtracking(tickets, ticket[1], used)if state: return Trueused[i] = Falseself.path.pop()  def findItinerary(self, tickets: List[List[str]]) -> List[str]:tickets.sort()used = [False] * len(tickets)self.backtracking(tickets, 'JFK', used)return self.result

python版本只有优化到 使用字典 并逆序查找才可以通过所有的测试用例。

from collections import defaultdictclass Solution:def findItinerary(self, tickets):targets = defaultdict(list)  # 创建默认字典，用于存储机场映射关系for ticket in tickets:targets[ticket[0]].append(ticket[1])  # 将机票输入到字典中for key in targets:targets[key].sort(reverse=True)  # 对到达机场列表进行字母逆序排序result = []self.backtracking("JFK", targets, result)  # 调用回溯函数开始搜索路径return result[::-1]  # 返回逆序的行程路径def backtracking(self, airport, targets, result):while targets[airport]:  # 当机场还有可到达的机场时next_airport = targets[airport].pop()  # 弹出下一个机场self.backtracking(next_airport, targets, result)  # 递归调用回溯函数进行深度优先搜索result.append(airport)  # 将当前机场添加到行程路径中

C++:

class Solution {
private:
unordered_map<string, map<string, int>> targets;
bool backtracking(int ticketNum, vector<string>& result) {if (result.size()==ticketNum+1) return true;for (pair<const string, int>& target: targets[result[result.size()-1]]) {if (target.second > 0) {result.push_back(target.first);target.second--;if (backtracking(ticketNum, result)) return true;result.pop_back();target.second++;}}return false;
}    public:   vector<string> findItinerary(vector<vector<string>>& tickets) {vector<string> result;result.push_back("JFK");for (const vector<string>& vec: tickets) {targets[vec[0]][vec[1]]++;}backtracking(tickets.size(), result);return result;}
};

51. N皇后（可跳过） 

代码随想录

视频讲解：这就是传说中的N皇后？ 回溯算法安排！| LeetCode：51.N皇后_哔哩哔哩_bilibili

Python:

class Solution:def __init__(self):self.result = []def isValid(self, row, col, chessboard):# 检查列for i in range(row):if chessboard[i][col] == 'Q':return False# 检查45度i, j = row-1, col-1while i>=0 and j>=0:if chessboard[i][j] == 'Q':return Falsei -= 1j-= 1# 检查135度i, j = row-1, col+1while i>=0 and j<len(chessboard):if chessboard[i][j] == 'Q':return Falsei -= 1j += 1return Truedef backtracking(self, n, row, chessboard):if row==n:self.result.append(chessboard[:])returnfor col in range(n):if self.isValid(row, col, chessboard):chessboard[row] = chessboard[row][:col] + 'Q' + chessboard[row][col+1:]self.backtracking(n, row+1, chessboard)chessboard[row] = chessboard[row][:col] + '.' + chessboard[row][col+1:]               returndef solveNQueens(self, n: int) -> List[List[str]]:if n==1: return [['Q']]    chessboard = ['.'*n for _ in range(n)]self.backtracking(n, 0, chessboard)return self.result

C++:

class Solution {
public:vector<vector<string>> result;void backtracking(int n, int row, vector<string>& chessboard) {if (row==n) {result.push_back(chessboard);return;}for (int col=0; col<n; col++) {if (isValid(row, col, chessboard)) {chessboard[row][col] = 'Q';backtracking(n, row+1, chessboard);chessboard[row][col] = '.';}}}bool isValid(int row, int col, vector<string>& chessboard) {// 检查列for (int i=0; i<row; i++) {if (chessboard[i][col]=='Q') return false;}// 检查45度for (int i=row-1, j = col-1; i>=0 && j>=0; i--, j--) {if (chessboard[i][j]=='Q') return false;}// 检查135度for (int i=row-1, j=col+1; i>=0 && j<chessboard.size(); i--, j++) {if (chessboard[i][j]=='Q') return false;}return true;}vector<vector<string>> solveNQueens(int n) {result.clear();std:: vector<std::string> chessboard(n, std::string(n, '.'));backtracking(n, 0, chessboard);return result;}
};

37. 解数独（可跳过） 

代码随想录

视频讲解：回溯算法二维递归？解数独不过如此！| LeetCode：37. 解数独_哔哩哔哩_bilibili

Python:

class Solution:def isValid(self, row, col, ele, board):# 行for i in range(9):if board[row][i] == ele: return False# 列for j in range(9):if board[j][col] == ele: return False# 九宫格start_row = (row//3) * 3start_col = (col//3) * 3for i in range(start_row, start_row+3):for j in range(start_col, start_col+3):if board[i][j] == ele: return Falsereturn Truedef backtracking(self, row, board):for row in range(9):for col in range(9):if board[row][col] != ".": continuefor ele in range(1, 10):if self.isValid(row, col, str(ele), board): board[row][col] = str(ele)if self.backtracking(row+1, board): return Trueboard[row][col] = '.'return Falsereturn Truedef solveSudoku(self, board: List[List[str]]) -> None:"""Do not return anything, modify board in-place instead."""self.backtracking(0, board)

C++:

class Solution {
public:bool isValid(int row, int col, char ele, vector<vector<char>>& board) {// 行for (int i=0; i<9; i++) {if (board[row][i]==ele) return false;}// 列for (int j=0; j<9; j++) {if (board[j][col]==ele) return false;}        // 九宫格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]==ele) return false;}}return true;}bool backtracking(vector<vector<char>>& board) {for (int row=0; row<9; row++) {for (int col=0; col<9; col++) {if (board[row][col]=='.') {for (char ele='1'; ele<='9'; ele++) {if (isValid(row, col, ele, board)) {board[row][col] = ele;if (backtracking(board)) return true;board[row][col] = '.';}                        }return false;}}}return true;}void solveSudoku(vector<vector<char>>& board) {backtracking(board);}
};

总结

代码随想录