代码随想录刷题第30天

明天就是大年三十了，首先祝各位朋友新年快乐，新春愉快！巧合的是，今天正好是回溯章节的收尾，这也是辞旧迎新的好兆头。

第一题是重新安排行程https://leetcode.cn/problems/reconstruct-itinerary/description/，一道难题，确实有点超出自己的能力就看看题解抄一遍代码ac了。相当于抄了一遍答案哈哈

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

第二题是大名鼎鼎的N皇后问题https://leetcode.cn/problems/n-queens/description/，直接上卡哥题解。思路并不难懂，通过每行遍历，将列依次放入皇后，向下一行递归，并调用isValid函数判断合法性，若当前位置合法，则放入皇后。

class Solution {
public:
    vector> result;
    void backtracking(int n, int row, vector& 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& chessboard, int n){
        for (int i = 0; i < n; 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 < n; i--, j++){
            if (chessboard[i][j] == 'Q') return false;
        }
        return true;
    }
    vector> solveNQueens(int n) {
    std::vector chessboard(n, std::string(n, '.'));
    backtracking(n, 0, chessboard);
    return result;
    }
};

第三题是解数独https://leetcode.cn/problems/sudoku-solver/，作为回溯章节的压轴题，与其他题目最大的不同就是该题需要两个for循环确定一个点。

class Solution {
public:
    bool backtracking(vector>& board){
        for (int i = 0; i < board.size(); i++){//行遍历
            for (int j = 0; j < board[0].size(); j++){//列遍历
                if (board[i][j] != '.') continue;
                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>& 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;
    }
    void solveSudoku(vector>& board) {
    backtracking(board);
    }
};

刷完了回溯，给我的感觉是回溯像一个已经有框架的楼房，只需要在某些部分上添加东西。