LeetCode 332 重新安排行程 2023.11.23
class Solution {
public:
//unordered_map<出发机场，map<到达机场, 航班次数>> targets
unordered_map<string, map<string, int>> targets;
//递归回溯遍历,ticketNum为一共的航班次数
bool backtracking(int ticketNum, vector<string>& result)
{
if(result.size() == ticketNum + 1)
return true;
//遍历各个由result[size-1]前往的目的地
for (pair<const string, int>& target : targets[result[result.size() - 1]])
// for (auto it = targets[result[result.size() - 1]].begin(); it != targets[result[result.size() - 1]].end(); ++it) {
// pair<const string, int>& target = *it;
// }
{
//当前到达点还没有走过
if(target.second > 0)
{
//加入答案中，到达该目的地
result.push_back(target.first);
//记录当前到达地已经走过
target.second--;
//当前到达点作为下一次遍历的出发点开始遍历
//如果返回了true则说明遍历完成，返回答案
if(backtracking(ticketNum, result))
return true;
//返回false，则回溯，遍历下一个到达点
result.pop_back();
target.second++;
}
}
return false;
}
vector<string> findItinerary(vector<vector<string>>& tickets) {
targets.clear();
//用于存储答案的变量
vector<string> result;
//记录从a地到达b地的航班次数
for(const vector<string>& vec : tickets)
targets[vec[0]][vec[1]]++;
//起始机场一定为JFK
result.push_back("JFK");
backtracking(tickets.size(), result);
return result;
}
};
LeetCode 51 N-皇后 2023.11.23
class Solution {
public:
//存储最后答案
vector<vector<string>> result;
//判断同一列、斜线上是否有皇后
bool isValid(int row, int col, vector<string>& chessboard, int n)
{
//检查同一列是否有有皇后
for (int i = 0; i < row; i++)
{
if(chessboard[i][col] == 'Q')
return false;
}
//检查45度角是否有皇后
for (int i = row - 1, j = col - 1; i >= 0 && j >= 0; i--, j--)
{
if(chessboard[i][j] == 'Q')
return false;
}
//检查135度角是否有皇后
for (int i = row - 1, j = col + 1; i >= 0 && j <= n; i--, j++)
{
if(chessboard[i][j] == 'Q')
return false;
}
return true;
}
void backtracking(int n, int row, vector<string>& chessboard)
{
if(row == n)
{
result.push_back(chessboard);
return;
}
for (int col = 0; col < n; col++)
{
//如果当前位置合法
if(isValid(row, col, chessboard, n))
{
chessboard[row][col] = 'Q';//放置皇后
backtracking(n, row+1, chessboard);
chessboard[row][col] = '.';//回溯，撤销皇后
}
}
}
vector<vector<string>> solveNQueens(int n) {
result.clear();
vector<string> chessboard(n, string(n, '.'));
backtracking(n, 0, chessboard);
return result;
}
};
LeetCode 37 解数独 2023.11.23
class Solution {
public:
bool isValid(int row, int col, char val, vector<vector<char>>& board)
{
//判断同一行是否有相同数字
for (int i = 0; i < 9; i++)
{
if(board[row][i] == val)
return false;
}
//判断同一列是否有相同数字
for (int i = 0; i < 9; i++)
{
if(board[i][col] == val)
return false;
}
int startRow = (row/3)*3;//开始行0（0,1,2）1(3,4,5)
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;
}
//递归回溯遍历
bool backtracking(vector<vector<char>> & board)
{
//遍历每行
for (int i = 0; i < board.size(); i++)
{
//遍历每列
for (int j = 0; j < board[0].size(); j++)
{
//在没有数的位置
if(board[i][j] == '.')
{
//遍历9个数
for (int k = '1'; k <= '9'; k++)
{
//判断是否合法
if(isValid(i, j, k, board))
{
board[i][j] = k;
// 如果找到合适一组立刻返回
if(backtracking(board))
return true;
// 回溯，撤销k
board[i][j] = '.';
}
}
//9个数都试完了，都不行，那么就返回false
return false;
}
}
}
return true;
}
void solveSudoku(vector<vector<char>>& board) {
backtracking(board);
}
};