【LeetCode】406. Queue Reconstruction by Height

问题描述

问题链接：https://leetcode.com/problems/queue-reconstruction-by-height/#/description

Suppose you have a random list of people standing in a queue. Each person is described by a pair of integers (h, k), where h is the height of the person and k is the number of people in front of this person who have a height greater than or equal to h. Write an algorithm to reconstruct the queue.

Note:

The number of people is less than 1,100.

Example

Input:[[7,0], [4,4], [7,1], [5,0], [6,1], [5,2]]Output:[[5,0], [7,0], [5,2], [6,1], [4,4], [7,1]]

我的代码

思路探索

通过观察后发现，如果先按照k进行分组，就可以用插入法得到答案。

写代码的过程非常曲折，最后发现问题出在一个变量的作用域弄错了，应该写在循环之外。

通过的代码

public class Solution {    private static class People {        int h;        int k;        public People(int h,int k){            this.h = h;            this.k = k;        }        public int[] toArray(){            int[] arr = new int[2];            arr[0] = h;            arr[1] = k;            return arr;        }    }    public int[][] reconstructQueue(int[][] people) {        /*        思路是这样的，首先根据k，把数据分成组，比如弄一个Map<Integer,List<People>>        然后从k为0的List开始，依次插队。        规则是这样的，拿到一个新元素开始，假设它的k=K。从list的最左边开始向右数，        找到第K+1个比它大或者跟它相等的元素，如果到了list末尾也停止。记住下标，然后将元素插入进去。        */        int len = people.length;        Map<Integer,ArrayList<People>> map = initMap(people);        // 开始排序        ArrayList<People> arr = new ArrayList<People>();        int curK = 0;        while(arr.size() != len || curK > 1000){            // int a = 0;            // if(a == 0){            //     throw new RuntimeException("fdfdf");            // }            ArrayList<People> list = map.get(curK);            curK++;            if(list == null){                continue;            }            for(People p : list){                insertToRightPlace(arr,p);            }        }        // 得到答案        int[][] result = new int[len][2];        for(int i = 0; i < len; i++){            result[i] = arr.get(i).toArray();        }        return result;    }    private void insertToRightPlace(ArrayList<People> list, People p){        int i = 0;        int count = 0;        for(; i < list.size(); i++){            if(list.get(i).h >= p.h){                count++;            }            if(count > p.k){ // 如果==的时候break了，会插入在这个元素的前面，导致不符合题意。                break;            }        }        list.add(i,p);    }    private Map<Integer,ArrayList<People>> initMap(int[][] people){        int len = people.length;        Map<Integer,ArrayList<People>> map = new HashMap<Integer,ArrayList<People>>();        for(int i = 0; i < len; i++){            int h = people[i][0];            int k = people[i][1];            ArrayList<People> list = map.get(k);            if(list == null){                list = new ArrayList<People>();                map.put(k,list);            }            list = map.get(k);            list.add(new People(h,k));        }        return map;    }}

打败了36.65的Java代码，也算不错了。好累啊。

讨论区

Easy concept with Python/C++/Java Solution

链接地址：https://discuss.leetcode.com/topic/60394/easy-concept-with-python-c-java-solution/2

嗯，在这个回答里学到了新的思路，不错不错。思路太长就不贴了。