LintCode领扣 题解丨微软面试题:大楼轮廓

简介: LintCode领扣 题解丨微软面试题:大楼轮廓

水平面上有 N 座大楼,每座大楼都是矩阵的形状,可以用一个三元组表示 (start, end, height),分别代表其在x轴上的起点,终点和高度。大楼之间从远处看可能会重叠,求出 N 座大楼的外轮廓线。

外轮廓线的表示方法为若干三元组,每个三元组包含三个数字 (start, end, height),代表这段轮廓的起始位置,终止位置和高度。

在线评测地址:https://www.lintcode.com/problem/the-skyline-problem/description?utm_source=sc-tc-sz

样例1

输入:
[

[1, 3, 3],
[2, 4, 4],
[5, 6, 1]

]
输出:
[

[1, 2, 3],
[2, 4, 4],
[5, 6, 1]

]
说明:
建筑物如下图所示,黄色部分表示建筑物

样例2

输入:
[

[1, 4, 3],
[6, 9, 5]

]
输出:
[

[1, 4, 3],
[6, 9, 5]

]
说明:
建筑物如下图所示,黄色部分表示建筑物

【题解】

使用九章算法强化班中讲过的 HashHeap 和扫描线算法。 Java 可以用 TreeSet / TreeMap, C++ 可以用 Map.

import java.util.*;

public class Solution {

class HashHeap {

ArrayList<Integer> heap;
String mode;
int size_t;
HashMap<Integer, Node> hash;

class Node {
  public Integer id;
  public Integer num;

  Node(Node now) {
    id = now.id;
    num = now.num;
  }

  Node(Integer first, Integer second) {

    this.id = first;
    this.num = second;
  }
}

public HashHeap(String mod) {
  // TODO Auto-generated constructor stub
  heap = new ArrayList<Integer>();
  mode = mod;
  hash = new HashMap<Integer, Node>();
  size_t = 0;
}

public int peek() {
  return heap.get(0);
}

public int size() {
  return size_t;
}

public Boolean isEmpty() {
  return (heap.size() == 0);
}

int parent(int id) {
  if (id == 0) {
    return -1;
  }
  return (id - 1) / 2;
}

int lson(int id) {
  return id * 2 + 1;
}

int rson(int id) {
  return id * 2 + 2;
}

boolean comparesmall(int a, int b) {
  if (a <= b) {
    if (mode == "min")
      return true;
    else
      return false;
  } else {
    if (mode == "min")
      return false;
    else
      return true;
  }

}

void swap(int idA, int idB) {
  int valA = heap.get(idA);
  int valB = heap.get(idB);

  int numA = hash.get(valA).num;
  int numB = hash.get(valB).num;
  hash.put(valB, new Node(idA, numB));
  hash.put(valA, new Node(idB, numA));
  heap.set(idA, valB);
  heap.set(idB, valA);
}

public Integer poll() {
  size_t--;
  Integer now = heap.get(0);
  Node hashnow = hash.get(now);
  if (hashnow.num == 1) {
    swap(0, heap.size() - 1);
    hash.remove(now);
    heap.remove(heap.size() - 1);
    if (heap.size() > 0) {
      siftdown(0);
    }
  } else {
    hash.put(now, new Node(0, hashnow.num - 1));
  }
  return now;
}

public void add(int now) {
  size_t++;
  if (hash.containsKey(now)) {
    Node hashnow = hash.get(now);
    hash.put(now, new Node(hashnow.id, hashnow.num + 1));

  } else {
    heap.add(now);
    hash.put(now, new Node(heap.size() - 1, 1));
  }

  siftup(heap.size() - 1);
}

public void delete(int now) {
  size_t--;
  Node hashnow = hash.get(now);
  int id = hashnow.id;
  int num = hashnow.num;
  if (hashnow.num == 1) {

    swap(id, heap.size() - 1);
    hash.remove(now);
    heap.remove(heap.size() - 1);
    if (heap.size() > id) {
      siftup(id);
      siftdown(id);
    }
  } else {
    hash.put(now, new Node(id, num - 1));
  }
}

void siftup(int id) {
  while (parent(id) > -1) {
    int parentId = parent(id);
    if (comparesmall(heap.get(parentId), heap.get(id)) == true) {
      break;
    } else {
      swap(id, parentId);
    }
    id = parentId;
  }
}

void siftdown(int id) {
  while (lson(id) < heap.size()) {
    int leftId = lson(id);
    int rightId = rson(id);
    int son;
    if (rightId >= heap.size()
        || (comparesmall(heap.get(leftId), heap.get(rightId)) == true)) {
      son = leftId;
    } else {
      son = rightId;
    }
    if (comparesmall(heap.get(id), heap.get(son)) == true) {
      break;
    } else {
      swap(id, son);
    }
    id = son;
  }
}

}

class Edge {

int pos;
int height;
boolean isStart;

public Edge(int pos, int height, boolean isStart) {
  this.pos = pos;
  this.height = height;
  this.isStart = isStart;
}

}

class EdgeComparator implements Comparator {

@Override
public int compare(Edge arg1, Edge arg2) {
  Edge l1 = (Edge) arg1;
  Edge l2 = (Edge) arg2;
  if (l1.pos != l2.pos)
    return compareInteger(l1.pos, l2.pos);
  if (l1.isStart && l2.isStart) {
    return compareInteger(l2.height, l1.height);
  }
  if (!l1.isStart && !l2.isStart) {
    return compareInteger(l1.height, l2.height);
  }
  return l1.isStart ? -1 : 1;
}

int compareInteger(int a, int b) {
  return a <= b ? -1 : 1;
}

}

List> output(List> res) {

List<List<Integer>> ans = new ArrayList<List<Integer>>();
if (res.size() > 0) {
  int pre = res.get(0).get(0);
  int height = res.get(0).get(1);
  for (int i = 1; i < res.size(); i++) {
    List<Integer> now = new ArrayList<Integer>();
    int id = res.get(i).get(0);
    if (height > 0) {
      now.add(pre);
      now.add(id);
      now.add(height);
      ans.add(now);
    }
    pre = id;
    height = res.get(i).get(1);
  }
}
return ans;

}

public List> buildingOutline(int[][] buildings) {

// write your code here
List<List<Integer>> res = new ArrayList<List<Integer>>();

if (buildings == null || buildings.length == 0
    || buildings[0].length == 0) {
  return res;
}
ArrayList<Edge> edges = new ArrayList<Edge>();
for (int[] building : buildings) {
  Edge startEdge = new Edge(building[0], building[2], true);
  edges.add(startEdge);
  Edge endEdge = new Edge(building[1], building[2], false);
  edges.add(endEdge);
}
Collections.sort(edges, new EdgeComparator());

HashHeap heap = new HashHeap("max");

List<Integer> now = null;
for (Edge edge : edges) {
  if (edge.isStart) {
    if (heap.isEmpty() || edge.height > heap.peek()) {
      now = new ArrayList<Integer>(Arrays.asList(edge.pos,
          edge.height));
      res.add(now);
    }
    heap.add(edge.height);
  } else {
    heap.delete(edge.height);
    if (heap.isEmpty() || edge.height > heap.peek()) {
      if (heap.isEmpty()) {
        now = new ArrayList<Integer>(Arrays.asList(edge.pos, 0));
      } else {
        now = new ArrayList<Integer>(Arrays.asList(edge.pos,
            heap.peek()));
      }
      res.add(now);
    }
  }
}
return output(res);

}

}
更多题解参见:https://www.jiuzhang.com/solution/the-skyline-problem/?utm_source=sc-tc-sz

相关文章
【微软面试题】请计算出1的个数
【微软面试题】请计算出1的个数
字节跳动------剑指offer专题精选谷歌、微软等知名IT企业典型面试题
字节跳动------剑指offer专题精选谷歌、微软等知名IT企业典型面试题
187 0
字节跳动------剑指offer专题精选谷歌、微软等知名IT企业典型面试题
|
机器学习/深度学习 算法
LintCode 题解丨微软面试题:寻找旋转排序数组中的最小值
LintCode 题解丨微软面试题:寻找旋转排序数组中的最小值
LintCode 题解丨微软面试题:寻找旋转排序数组中的最小值
LintCode领扣 题解丨 谷歌面试题:二叉树的序列化和反序列化
LintCode领扣 题解丨 谷歌面试题:二叉树的序列化和反序列化
LintCode领扣 题解丨 谷歌面试题:二叉树的序列化和反序列化
LintCode领扣 题解丨 美团面试真题:二叉查找树的中序后继
LintCode领扣 题解丨 美团面试真题:二叉查找树的中序后继
LintCode领扣 题解丨 美团面试真题:二叉查找树的中序后继
LintCode领扣 题解丨腾讯面试题:被围绕的区域
LintCode领扣 题解丨腾讯面试题:被围绕的区域
LintCode领扣 题解丨腾讯面试题:被围绕的区域
LintCode领扣 题解丨 BAT大厂面试高频题:逆波兰表达式求值
LintCode领扣 题解丨 BAT大厂面试高频题:逆波兰表达式求值
LintCode领扣 题解丨 BAT大厂面试高频题:逆波兰表达式求值
|
机器学习/深度学习 算法 索引
LintCode领扣 题解丨 谷歌面试高频题:最大假期天数
LintCode领扣 题解丨 谷歌面试高频题:最大假期天数
LintCode领扣 题解丨 谷歌面试高频题:最大假期天数