【leetcode】【M】406. Queue Reconstruction by Height【95】

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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贪心。

最开始，做法是，按照个子大小排序，从小到大

然后一个一个安排，判断是不是要放到res里面

判断的方法是，判断整个res里面有几个能挡住他的，如果个数跟k相等，那就放进去

整个过程循环进行， 直到p里面一个元素都没有了

然后，结果虽然对，超时了。

找人家的算法。

想象一下排座位，h是身高，k是某个小朋友最多允许几个人挡住他（貌似不是很恰当）。

对于单个小朋友来说，只有比他高才会挡住他，所以我们先按h的标准排最高的一批小朋友，这样一来后面批次的小朋友不会影响自己，因为后来比自己矮的不可能挡住自己。

[[7,0], [4,4], [7,1], [5,0], [6,1], [5,2]]

[7,0]和[7,1]先排好。

同一批次里，K小的在前面，这个很好理解。。

第二批小朋友是身高为6的，[6,1]，他要保证只有1个人挡住他，他坐到刚才2个人之间就行了。

[7,0] [6,1] [7,1]

第三批是身高为5的，[5,0]和[5,2]，[5,0]想近距离观察美女老师，然而，目前在坐的所有人都比他高（我不是针对谁，还没坐下的都是勒涩），他就跑去第一排了。至于[5,2]，性欲不那么旺盛，被2个人挡住也无所谓，于是坐到2个人后面。

[5,0] [7,0] [6,1] [7,1]

[5,0] [7,0] [5,2] [6,1] [7,1]

然后是最后一个小朋友[4,4]，他可能比较像我，坐怀不乱，正人君子，道德楷模，上课的时候心无旁骛，不需要看美女老师，所以前面4个人无所谓，他跑到4个人后面

[5,0] [7,0] [5,2] [6,1] [4,4] [7,1]

BB半天说白了一句话，h大的前面，h一样的k小的在前面。
根据这句话设立PQ的判断规则就行了。。

你看，说的很明白了，显然这个算法更快也更简洁

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 xrange(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

# for i in peopledct.items():
#     print i

# sort from the tallest group, to ensure that the tallest is always in the front
for h in height[::-1]:
peopledct[h].sort()
for p in peopledct[h]:
res.insert(p[0], people[p[1]])
# p[0] mean the number of person higher than him on front of him
# print res

return res