406. Queue Reconstruction by Height (Medium)
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原题目:
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.
ExampleInput:[[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]]
题目大意如下:
给定一个随机排序人的队列,队列里面的人用[h,k]二元组表示,h代表当前人的高度,k代表在这个人前面且高度大于等于它的h的人数,现在要求按照这个二元组重构这个队列使之变得有效。
解题思路:
还是排序,先按照他们的高度从高到低(因为我后面用的从前往后遍历插入,当然也可以从低到高)排序,如果高度一样,那么按照k值从小到大排序。
排完序后我们可以注意到这样一个事实:如果我们按照先处理身最高的,那他们的k值就是他们所应该在的位置——因为已经没有比他们更高的了。
所以我们从高度从高到低按照k值的位置一直插入到答案中即可。
代码如下:
class Solution {public: static bool cmp (pair<int, int> a , pair<int, int> b){ if(a.first == b.first ) return a.second < b.second ; return a.first > b.first ; } vector<pair<int, int>> reconstructQueue(vector<pair<int, int>>& people) { vector<pair<int, int>> ans ; ans.clear() ; sort(people.begin() , people.end() , cmp) ; for(auto i : people) //此时前面的人一定都比后面的高所以他们的k就是他们应该在的位置 ans.insert( ans.begin() + i.second , i) ; return ans ; }};
运行结果:
0 0
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