文章目录
- 重新安排行程
- 题解
- [N 皇后](https://leetcode.cn/problems/n-queens/description/)
- 题解
- 解数独
- 题解
重新安排行程
| Category | Difficulty | Likes | Dislikes | ContestSlug | ProblemIndex | Score |
|---|---|---|---|---|---|---|
| algorithms | Hard (47.57%) | 765 | 0 | - | - | 0 |
给你一份航线列表 tickets ，其中 tickets[i] = [fromi, toi] 表示飞机出发和降落的机场地点。请你对该行程进行重新规划排序。
所有这些机票都属于一个从 JFK（肯尼迪国际机场）出发的先生，所以该行程必须从 JFK 开始。如果存在多种有效的行程，请你按字典排序返回最小的行程组合。
- 例如，行程
["JFK", "LGA"]与["JFK", "LGB"]相比就更小，排序更靠前。
假定所有机票至少存在一种合理的行程。且所有的机票 必须都用一次 且 只能用一次。
示例 1：
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输入：tickets = [["MUC","LHR"],["JFK","MUC"],["SFO","SJC"],["LHR","SFO"]]
输出：["JFK","MUC","LHR","SFO","SJC"]
示例 2：
输入：tickets = [["JFK","SFO"],["JFK","ATL"],["SFO","ATL"],["ATL","JFK"],["ATL","SFO"]]
输出：["JFK","ATL","JFK","SFO","ATL","SFO"]
解释：另一种有效的行程是 ["JFK","SFO","ATL","JFK","ATL","SFO"] ，但是它字典排序更大更靠后。
提示：
1 <= tickets.length <= 300tickets[i].length == 2fromi.length == 3toi.length == 3fromi和toi由大写英文字母组成fromi != toi
Discussion | Solution
题解
// @lc code=start
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) {
targets.clear();
vector<string> result;
for(const vector<string>& vec : tickets) {
targets[vec[0]][vec[1]]++;
}
result.push_back("JFK");
backtracking(tickets.size(),result);
return result;
}
};
[参考文章](代码随想录 (programmercarl.com))
N 皇后
| Category | Difficulty | Likes | Dislikes | ContestSlug | ProblemIndex | Score |
|---|---|---|---|---|---|---|
| algorithms | Hard (74.17%) | 1726 | 0 | - | - | 0 |
按照国际象棋的规则，皇后可以攻击与之处在同一行或同一列或同一斜线上的棋子。
n 皇后问题 研究的是如何将 n 个皇后放置在 n×n 的棋盘上，并且使皇后彼此之间不能相互攻击。
给你一个整数 n ，返回所有不同的 n 皇后问题 的解决方案。
每一种解法包含一个不同的 n 皇后问题 的棋子放置方案，该方案中 'Q' 和 '.' 分别代表了皇后和空位。
示例 1：
输入：n = 4
输出：[[".Q..","...Q","Q...","..Q."],["..Q.","Q...","...Q",".Q.."]]
解释：如上图所示，4 皇后问题存在两个不同的解法。
示例 2：
输入：n = 1
输出：[["Q"]]
提示：
1 <= n <= 9
Discussion | Solution
题解
// @lc code=start
class Solution {
private:
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,n)) {
chessboard[row][col] = 'Q';
backtracking(n,row + 1,chessboard);
chessboard[row][col] = '.';
}
}
}
bool isValid(int row,int col,vector<string>& chessboard, int n) {
for(int i = 0; i < row; ++i) {
if(chessboard[i][col] == 'Q') {
return false;
}
}
for(int i = row - 1, j = col - 1; i >= 0 && j >= 0 ; i--,j--) {
if(chessboard[i][j] == 'Q') {
return false;
}
}
for(int i = row - 1, j = col + 1; i >= 0 && j >=0; i--, j++) {
if(chessboard[i][j] == 'Q') {
return false;
}
}
return true;
}
public:
vector<vector<string>> solveNQueens(int n) {
result.clear();
vector<string> chessboard(n,string(n,'.'));
backtracking(n,0,chessboard);
return result;
}
};
[参考文章](代码随想录 (programmercarl.com))
解数独
| Category | Difficulty | Likes | Dislikes | ContestSlug | ProblemIndex | Score |
|---|---|---|---|---|---|---|
| algorithms | Hard (67.60%) | 1596 | 0 | - | - | 0 |
数组 | 回溯 | 矩阵
编写一个程序，通过填充空格来解决数独问题。
数独的解法需 遵循如下规则：
- 数字
1-9在每一行只能出现一次。 - 数字
1-9在每一列只能出现一次。 - 数字
1-9在每一个以粗实线分隔的3x3宫内只能出现一次。（请参考示例图）
数独部分空格内已填入了数字，空白格用 '.' 表示。
示例 1：
输入：board = [["5","3",".",".","7",".",".",".","."],["6",".",".","1","9","5",".",".","."],[".","9","8",".",".",".",".","6","."],["8",".",".",".","6",".",".",".","3"],["4",".",".","8",".","3",".",".","1"],["7",".",".",".","2",".",".",".","6"],[".","6",".",".",".",".","2","8","."],[".",".",".","4","1","9",".",".","5"],[".",".",".",".","8",".",".","7","9"]]
输出：[["5","3","4","6","7","8","9","1","2"],["6","7","2","1","9","5","3","4","8"],["1","9","8","3","4","2","5","6","7"],["8","5","9","7","6","1","4","2","3"],["4","2","6","8","5","3","7","9","1"],["7","1","3","9","2","4","8","5","6"],["9","6","1","5","3","7","2","8","4"],["2","8","7","4","1","9","6","3","5"],["3","4","5","2","8","6","1","7","9"]]
解释：输入的数独如上图所示，唯一有效的解决方案如下所示：
提示：
board.length == 9board[i].length == 9board[i][j]是一位数字或者'.'- 题目数据 保证 输入数独仅有一个解
Discussion | Solution
题解
// @lc code=start
class Solution {
private:
bool backtracking(vector<vector<char>>& board) {
for(int i = 0; i < board.size(); i++) {
for(int j = 0; j < board[0].size();j++) {
if(board[i][j] == '.') {
for(char k = '1'; k <= '9';k++) {
if(isValid(i,j,k,board)) {
board[i][j] = k;
if(backtracking(board)) return true;
board[i][j] = '.';
}
}
return false;
}
}
}
return true;
}
bool isValid(int row,int col,char val,vector<vector<char>>& board) {
for(int i = 0; i < 9;i++) {
if(board[row][i] == val) {
return false;
}
}
for(int j = 0; j < 9; j++) {
if(board[j][col] == val) {
return false;
}
}
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] == val) {
return false;
}
}
}
return true;
}
public:
void solveSudoku(vector<vector<char>>& board) {
backtracking(board);
}
};
startCol; j < startCol + 3; j++) {
if(board[i][j] == val) {
return false;
}
}
}
return true;
}
public:
void solveSudoku(vector<vector>& board) {
backtracking(board);
}
};
[参考文章]([代码随想录 (programmercarl.com)](https://programmercarl.com/0037.解数独.html#总结))
