华为OD面试题

题目

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思路

方法一

- B座为空，判断A座、C座是否为空，一个座为空，TicketNums++
- D座为空，判断F座是否为空，为空，TicketNums++
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方法二

当所有座位均可选时：TicketNums = rows * 2
- B座不为空，TicketNums--
- A座不为空并且B座为空，判断C座是否为空，不为空，TicketNums--
- C座不为空并且B座为空，判断A座是否为空，不为空，TicketNums--
- D座不为空并且F座为空，TicketNums--
- F座不为空并且D座为空，TicketNums--
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实现

方法一

    public int getTicketNumsFew(int rows, List<String> seats){

        int[][] Seats = new int[rows+1][6];

        int num = 0;

        for (String seat : seats) {

            String[] chars = seat.split("");

            int row = Integer.parseInt(chars[0]);

            switch (chars[1]){
                case "A":
                    Seats[row][1] = 1;
                    break;
                case "B":
                    Seats[row][2] = 1;
                    break;
                case "C":
                    Seats[row][3] = 1;
                    break;
                case "D":
                    Seats[row][4] = 1;
                    break;
                case "F":
                    Seats[row][5] = 1;
                    break;
            }
        }

        for (int i = 1; i <= rows; i++){
            if (Seats[i][2] != 1 && (Seats[i][1] != 1 || Seats[i][3] != 1)){
                num++;
            }
            if (Seats[i][4] != 1 && Seats[i][5] != 1){
                num++;
            }
        }
        
        return num;
    }
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方法二

    public int getTicketNumsLarge(int rows, List<String> seats){

        int[][] Seats = new int[rows+1][6];

        int num = rows * 2;

        for (String seat : seats) {

            String[] chars = seat.split("");

            int row = Integer.parseInt(chars[0]);

            switch (chars[1]){
                case "A":
                    Seats[row][1] = 1;
                    if (Seats[row][2] != 1 && Seats[row][3] == 1){
                        num--;
                    }
                    break;
                case "B":
                    Seats[row][2] = 1;
                    num--;
                    break;
                case "C":
                    Seats[row][3] = 1;
                    if (Seats[row][2] != 1 && Seats[row][1] == 1){
                        num--;
                    }
                    break;
                case "D":
                    Seats[row][4] = 1;
                    if (Seats[row][5] != 1){
                        num--;
                    }
                    break;
                case "F":
                    Seats[row][5] = 1;
                    if (Seats[row][4] != 1){
                        num--;
                    }
                    break;
            }
        }

        return num;
    }
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原文地址：TT的博客《华为OD面试题》