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字节跳动的三道编码面试题的实现

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国庆节后,自己的一个小圈子微信群的伙伴们发了一张图片,是网上流传的字节跳动的面试题编码,闲的无事就思索了下,发现都不难,都是对基础的数学知识的考量。先上图吧!

当然40分钟,我也无法把任意两题编码完成,只是知道大概的解题思路,唯一能确定的,在面试规定时间内,第二题我是肯定可以在20分钟内编码完成。

题目一

基础知识就是初中的平面直角坐标系,解析思路:

  1. 计算总周长;
  2. 将各边长的前后坐标计算出来封装好,第四步要使用;
  3. 根据K段值计算出平均分段后的长度;
  4. 然后循环K次,根据平均长度依次相加计算等分点的坐标。

不多说,上代码:

先定义坐标的Point类

class Point {
        float x;
        float y;

        public Point() {
        }

        public Point(float x, float y) {
            this.x = x;
            this.y = y;
        }

        public Point(Point point) {
            this(point.x, point.y);
        }

        @Override
        public String toString() {
            return "Point, x:" + x + " y:" + y;
        }
    }

N边形的边封装类

class Line {
        Point begin;
        Point end;
        float length;

        public Line() {

        }

        public Line(Point begin, Point end, float length) {
            this.begin = begin;
            this.end = end;
            this.length = length;
        }
    }

现在上实现计算的类

这段代码第一个版本的时候,在正方形偶数等分的时候,坐标点计算不准确,今晚上看着代码思考了10分钟的样子,稍微改动了下,暂时没有这个bug了。其他的bug,期待大家一起发现,然后修复吧!

public class Polygon {

    /**
     * 计算边的长度
     * 
     * @return
     */
    private static float lineLength(Point a, Point b) {
        float length;

        if (a.x == b.x) {
            // 垂直线条
            length = Math.abs(a.y - b.y);
        } else {
            length = Math.abs(a.x - b.x);
        }

        return length;
    }

    /**
     * 计算 周长
     * 
     * @return
     */
    private static float totalSideLength(Point[] points, Line[] lines) {
        float side = 0;

        for (int i = 1; i < points.length; i++) {
            Point prev = points[i - 1];
            Point point = points[i];

            float length = lineLength(prev, point);

            side += length;
            lines[i - 1] = new Line(prev, point, length);

            if (i == points.length - 1) {
                length = lineLength(point, points[0]);

                side += length;
                lines[i] = new Line(point, points[0], length);
            }
        }

        return side;
    }

    public static Point[] division(Point[] points, int divisionNum) {
        Point[] divisionPoint = new Point[divisionNum];

        // 计算周长
        Line[] lines = new Line[points.length];
        float side = totalSideLength(points, lines);

        // 等分长度
        float divisionLength = side / divisionNum;

        int lineIndex = -1;
        float sumLength = 0;

        for (int i = 0; i < divisionNum; i++) {
            if (i == 0) {
                // 第一个等分点直接是起始点坐标
                divisionPoint[i] = new Point(points[0]);
                continue;
            }

            divisionPoint[i] = new Point();
            float lineLength = divisionLength * i;

            while (true) {
                Line line;
                if (sumLength < lineLength) {
                    lineIndex++;
                    line = lines[lineIndex];
                    sumLength += line.length;
                } else
                    line = lines[lineIndex];

                if (sumLength >= lineLength) {
                    float temp = sumLength - lineLength;

                    if (line.begin.x == line.end.x) {
                        // begin和end的坐标点垂直
                        divisionPoint[i].x = line.begin.x;

                        if (line.end.y > line.begin.y)
                            divisionPoint[i].y = line.end.y - temp;
                        else
                            divisionPoint[i].y = line.end.y + temp;
                    } else {
                        // begin和end的坐标点水平
                        divisionPoint[i].y = line.end.y;

                        if (line.end.x > line.begin.x)
                            divisionPoint[i].x = line.end.x - temp;
                        else
                            divisionPoint[i].x = line.end.x + temp;
                    }
                    
                    break;
                }
            }
        }

        return divisionPoint;
    }

    private static void print(Point[] points) {
        for (int i = 0; i < points.length; i++) {
            System.out.println("第" + (i + 1) + "等分点, x:" + points[i].x + ",y:" + points[i].y);
        }
    }

    public static void main(String[] args) {
        Point[] points = new Point[] { new Point(0, 0), new Point(0, 1), new Point(1, 1), new Point(1, 0) };

        Point[] divPoints = division(points, 8);

        print(divPoints);
    }
}

题目二

解题思路:

对应位数的数字相加,永远不会超过18,所以,我们就先把对应位置的和计算出来,然后再反复循环找到大于9的数,向高位进位。

这个比较简单,只是考察个位数的正整数加法永远不大于18这个细节。

上代码:

public class LinkAddition {
    static class NumNode {
        public int num;
        public NumNode next;

        public NumNode() {
        }

        public NumNode(int num) {
            this.num = num;
        };

        public NumNode(int num, NumNode next) {
            this(num);
            this.next = next;
        }
    }

    private static int length(NumNode num) {
        int length = 0;

        NumNode temp = num;
        while (temp != null) {
            length++;
            temp = temp.next;
        }

        return length;
    }

    private static NumNode calc(NumNode a, NumNode b, int aLength, int bLength) {
        NumNode aNode = a;
        NumNode bNode = b;

        NumNode result = new NumNode();
        NumNode resultNode = result;

        // 计算b链表再a中的起始索引
        int aStartIndex = aLength - bLength;

        for (int i = 0; i < aLength; i++) {
            if (i >= aStartIndex) {
                resultNode.num = aNode.num + bNode.num;
                bNode = bNode.next;
            } else
                resultNode.num = aNode.num;

            aNode = aNode.next;
            if (aNode != null) {
                resultNode.next = new NumNode();
                resultNode = resultNode.next;
            }
        }

        return result;
    }

    public static NumNode addition(NumNode a, NumNode b) {
        NumNode result = null;

        // 计算位数
        int aLength = length(a);
        int bLength = length(b);

        if (aLength > bLength) {
            result = calc(a, b, aLength, bLength);
        } else {
            result = calc(b, a, bLength, aLength);
        }

        boolean isGreater9 = true;

        while (isGreater9) {
            isGreater9 = false;
            NumNode node = result;

            while (node != null) {
                // 检查是否有大于9的节点
                if (node.num > 9) {
                    isGreater9 = true;
                    break;
                }

                node = node.next;
            }

            // 没有大于9且需要进位的节点
            if (!isGreater9)
                break;
            
            node = result;
            
            if (node.num > 9) {
                // 头节点的内容跟大于9,需要进位
                result = new NumNode(1, node);

                node.num = node.num - 10;
            }

            while (node.next != null) {
                if (node.next.num > 9) {
                    node.num += 1;
                    node.next.num = node.next.num - 10;
                }
                node = node.next;
            }
        }

        return result;
    }

    private static void print(NumNode num) {
        NumNode node = num;
        while (node != null) {
            System.out.print(node.num);
            node = node.next;
        }
    }

    public static void main(String[] args) {
        NumNode a = new NumNode(9);
        a.next = new NumNode(9, new NumNode(9));

        NumNode b = new NumNode(9);
        // b.next = new NumNode(9, new NumNode(9));

        NumNode result = addition(a, b);

        print(result);
    }
}

题目三

这个我写的第一个版本,只契合类那个举例,然后瞬间就被我推翻类,最后坐下思考类10分钟,把这个按照二维数组的思路解析了。

先找到最高处,然后就以最高处为一个维度,做循环计算出水量,还是上代码吧:

public class Water {
    public static int waterNum(int[] steps) {
        int waterNum = 0;

        int max = steps[0];
        for (int i = 1; i < steps.length; i++) {
            if (max < steps[i])
                max = steps[i];
        }

        for (int i = 0; i < max; i++) {
            int num = 0, index = 0;

            for (int n = 0; n < steps.length; n++) {
                if (steps[n] - i > 0) {
                    if (num > 0) {
                        waterNum += n - index - 1;
                    }

                    num = steps[n] - i;
                    index = n;
                }
            }
        }

        return waterNum;
    }

    public static void main(String[] args) {
        int[] steps = new int[] { 0, 1, 0, 2, 1, 0, 1, 3, 2, 1, 2, 3, 0, 1 };
        int water = waterNum(steps);

        System.out.println(water);
    }
}

总结:

其实这几题本身的知识点并不难,都是平时用到的,就看怎么转化为代码罢了。

第一题考察的直角坐标系上怎么计算边长,然后根据均分等长从第一条边挨着走,计算对应的坐标,该知识点在初中就已学过。

第二题则是考察每位上的正整数加法到底最大能到多少,只要明白了这一点,把每一位上相加后,再统一做进位处理就可以了。

第三题的代码量是最少的,我的解题思路是二位数组的方式, 也不算难。

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