代码随想录一刷day30

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该文介绍了三道力扣（LeetCode）上的算法问题，包括332.重新安排行程，使用深度优先搜索和回溯法解决航班路径规划；51.N皇后问题，利用回溯法在棋盘上放置皇后避免冲突；以及37.解数独，通过递归和九宫格规则填充值。

一、力扣332.重新安排行程

class Solution {
    private LinkedList<String> res;
    private LinkedList<String> path = new LinkedList<>();
    public List<String> findItinerary(List<List<String>> tickets) {
        Collections.sort(tickets, (a, b) -> a.get(1).compareTo(b.get(1)));
        path.add("JFK");
        boolean[] used = new boolean[tickets.size()];
        backTracking((ArrayList) tickets, used);
        return res;
    }
    public boolean backTracking(ArrayList<List<String>> tickets, boolean[] used) {
        if (path.size() == tickets.size() + 1) {
            res = new LinkedList(path);
            return true;
        }

        for (int i = 0; i < tickets.size(); i++) {
            if (!used[i] && tickets.get(i).get(0).equals(path.getLast())) {
                path.add(tickets.get(i).get(1));
                used[i] = true;

                if (backTracking(tickets, used)) {
                    return true;
                }

                used[i] = false;
                path.removeLast();
            }
        }
        return false;
    }
}

二、力扣51. N 皇后

class Solution {
    private List<List<String>> res = new ArrayList<>();
    public List<List<String>> solveNQueens(int n) {
        char[][] chessboard  = new char[n][n];
        for(char[] ch : chessboard){
            Arrays.fill(ch, '.');
        }
        fun(n, 0, chessboard);
        return res;
    }
    private void fun(int n, int row, char[][] chessboard){
        if(row == n){
            res.add(arrayToStringFun(chessboard));
            return ;
        }
        for(int i = 0; i < n; i ++){
            if(isFlag(chessboard, n, row, i)){
                chessboard[row][i] = 'Q';
                fun(n, row + 1, chessboard);
                chessboard[row][i] = '.';
            }
        }
    }
    private List<String> arrayToStringFun(char[][] chessboard){
        List<String> path = new ArrayList<>();
        for(char[] ch : chessboard){
            path.add(String.copyValueOf(ch));
        }
        return path;
    }
    private boolean isFlag(char[][] chessboard, int n, int row, int col){
        //判断列
        for(int i = 0; i < n; i ++){
            if(chessboard[i][col] == 'Q'){
                return false;
            }
        }
        //判断45度
        for(int i = row-1, j = col-1; i >= 0 && j >= 0; i--, j--){
            if(chessboard[i][j] == 'Q'){
                return false;
            }
        }
        //判断135度
        for(int i = row-1, j = col+1; i >= 0 && j < n; i--, j++){
            if(chessboard[i][j] == 'Q'){
                return false;
            }
        }
        return true;
    }
}

三、力扣37. 解数独

class Solution {
    public void solveSudoku(char[][] board) {
        fun(board);
    }
    private boolean fun(char[][] board){
        for(int i = 0; i < 9; i ++){
            for(int j = 0; j < 9; j ++){
                if(board[i][j] != '.'){
                    continue;
                }
                for(char c = '1'; c <= '9'; c ++){
                    if(isfun(i, j, c, board)){
                        board[i][j] = c;
                        if(fun(board)){
                            return true;
                        }else{
                            board[i][j] = '.';
                        }
                    }
                }
                return false;
            }
        }
        return true;
    }
    private boolean isfun(int i, int j, char c, char[][] board){
        //同行是否重复
        for(int a = 0; a < 9 ; a++){
            if(board[i][a] == c){
                return false;
            }
        }
        //同列是否重复
        for(int b = 0; b < 9; b ++){
            if(board[b][j] == c){
                return false;
            }
        }
        //九宫格内是否合规
        int startrRow = (i/3)*3;
        int startrCol = (j/3)*3;
        for(int a = startrRow; a < startrRow+3; a ++){
            for(int b = startrCol; b < startrCol + 3; b ++){
                if(board[a][b] == c){
                    return false;
                }
            }
        }
        return true;
    }
}