python--leetcode406. 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]]

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本题的题目意思是给你一个随机排序的list，list中每个元素为(h,k)。h为此人身高，k为身高大于等于此人的人数。请写一个算法重构这个队列。

看上去比较复杂，所以得细细讲一下解题思路：

步骤如下：

1、挑出最高的一群人，并在一个子列表（S）中排序。 既然没有其他人比他们高，所以每个人的下标就和他的k值一样。
2、对于第二个最高的组，将它们中的每一个以k值为下标插入（S）。之后都以此类推。
例如。
输入：[[7,0]，[4,4]，[7,1]，[5,0]，[6,1]，[5,2]]
步骤1之后的子阵列：[[7,0]，[7,1]]
步骤2之后的子阵列：[[7,0]，[6,1]，[7,1]]
...


看代码：

class Solution(object):    def reconstructQueue(self, people):        if not people: return []        # obtain everyone's info        # key=height, value=k-value, index in original array        peopledct, height, res = {}, [], []        for i in range(len(people)):            p = people[i]            if p[0] in peopledct:                peopledct[p[0]] += (p[1], i),            else:                peopledct[p[0]] = [(p[1], i)]                height += p[0],        height.sort()  # here are different heights we have        # sort from the tallest group        for h in height[::-1]:            peopledct[h].sort()            for p in peopledct[h]:                # 此时p[0]为插入位置，people[p[1]]为要插入的元素                res.insert(p[0], people[p[1]])        return ress=Solution()print(s.reconstructQueue([[7,0], [4,4], [7,1], [5,0], [6,1], [5,2]]))