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  • 406. Queue Reconstruction by Height

    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]]

    refer to: https://discuss.leetcode.com/topic/60394/easy-concept-with-python-c-java-solution

    1. Pick out tallest group of people and sort them in a subarray (S). Since there's no other groups of people taller than them, therefore each guy's index will be just as same as his k value.
    2. For 2nd tallest group (and the rest), insert each one of them into (S) by k value. So on and so forth.

    E.g.
    input: [[7,0], [4,4], [7,1], [5,0], [6,1], [5,2]]
    subarray after step 1: [[7,0], [7,1]]
    subarray after step 2: [[7,0], [6,1], [7,1]]

    
    
      
     
    public class Solution {
        public int[][] reconstructQueue(int[][] people) {
            //pick up the tallest guy first
            //when insert the next tall guy, just need to insert him into kth position
            //repeat until all people are inserted into list
            Arrays.sort(people,new Comparator<int[]>(){
               @Override
               public int compare(int[] o1, int[] o2){
                   return o1[0]!=o2[0]? o2[0]-o1[0] : o1[1]-o2[1];
               }
            });
            List<int[]> res = new LinkedList<>();
            for(int[] cur : people){
                res.add(cur[1],cur);       
            }
            return res.toArray(new int[people.length][]);
        }
    }`
    
    res.add(cur[1],cur); linkedList 在指定位置插入  toArray(new Object[0]) 和 toArray() 在功能上是相同的。
    所以要用res.toArray(new int[people.length][]);来表示[][]数组啊

     

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  • 原文地址:https://www.cnblogs.com/apanda009/p/7148196.html
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