[leetcode] 406. Queue Reconstruction by Height 解题报告
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题目链接:https://leetcode.com/problems/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]]
思路:一开始的思路是按照k的大小升序排列,如果k相等就按h升序排列.然后对于每一个元素从头开始到自身位置查找并计数比当前元素大的个数,直到计数比当前元素k大停止,如果此时正好停在当前位置,说明当前元素不需要移动,否则就删除当前元素插入到停止位置之前.这样做的好处是不需要额外申请空间,但是代码稍微麻烦一点.看了一下别人的,发现只需要改变一下排序的规则可以让代码简化很多.就是依然按照h降序排序,但是如果h相等则按照k升序排序.然后开一个新数组,因为数组按照降序排序,所以每次只需要将元素插入到新数组的k的位置即可.两种方法时间复杂度都是O(n*2),但是第二种更容易理解和实现.
代码如下:
class Solution {public: vector<pair<int, int>> reconstructQueue(vector<pair<int, int>>& people) { if(people.size()==0) return {}; auto cmp = [](pair<int, int> a, pair<int, int> b) { return a.first==b.first?a.second<b.second:a.first>b.first; }; sort(people.begin(), people.end(), cmp); vector<pair<int, int>> ans; for(auto val: people) ans.insert(ans.begin() + val.second, val); return ans; }};
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