[leetcode] 406. Queue Reconstruction by Height
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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]]
这道题是重塑队列,题目难度为Medium。
根据题目描述,我们首先选出高度最高的一批人,把他们按h由小到大排序,他们之间的相对顺序就确定了;之后选出次高的一批人,然后根据他们的h值将其插入现有新队列中;如此迭代直至原队列中所有人都插入新队列。拿题目例子说明:首先选出最高的一批人[7,0]、[7,1],排序后插入队列,当前队列为[[7,0],[7,1]];然后选出下一批人[6,1],由于其h值为1,所以在队列中下标为1的位置插入[6,1],当前队列为[[7,0],[6,1],[7,1]];再选出下一批人[5,0]、[5,2],根据各自h值插入队列后队列为[[5,0],[7,0],[5,2],[6,1],[7,1]];最后选出[4,4],根据h值插入相应位置得到最终队列[[5,0],[7,0],[5,2],[6,1],[4,4],[7,1]]。
之所以按此思路重塑队列,是因为某高度的人在插入队列时不会影响比他高的人,因此按高度由高到低的顺序分批次将所有人插入队列即可。这样题目就变成了排序问题,按照以上描述的思路排序之后顺次插入队列即可完成队列重塑。具体代码:
class Solution {public: vector<pair<int, int>> reconstructQueue(vector<pair<int, int>>& people) { vector<pair<int, int>> ret; sort(people.begin(), people.end(), [](pair<int, int> l, pair<int, int> r) { return (l.first > r.first) || (l.first == r.first && l.second < r.second); }); for(auto p:people) ret.insert(ret.begin()+p.second, p); return ret; }};
另外,在vector上执行插入操作效率比较低,如果换为list则没有随机访问迭代器,需要根据h值遍历到插入位置,但插入效率较高,大家可以比较下两者的差别。
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