题解——Leetcode 406. Queue Reconstruction by Height 难度:Medium
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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 toh
. Write an algorithm to reconstruct the queue.
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]]
题目如上,简化后的题意如下:
给你一个队列,里面是随机排列的序对。每一个序对表示一个人,序对的第一个元素表示身高,第二个元素表示队列前面不低于自己身高的人数。要求按照序对的含义重新排列队列。
C++程序如下:
class Solution {public: static bool compare(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>> result; if(people.size() == 0) return result; sort(people.begin(), people.end(), compare); for(int i = 0; i < people.size(); i++){ result.insert(result.begin() + people[i].second, people[i]); } return result; }};将具有最高身高的人先挑出来组成一个子队列,如果最高的人不止一个,按照第二个元素的升序排列。接着挑出第二高的人,同样按照第二个元素的升序排列。由于比他们高的人都已经挑出来了,所以第二个元素的值即为其在队列中应处的位置(队列下标从0开始),将第二高的人插入之前的子队列形成一个更长的队列。重复以上过程直到所有人都插进了前一步骤的子队列,插入完成后的队列即为符合题意的新队列。
compare()函数把队列按照序对第一个元素的降序排列,第一个元素相同的序对按照第二个元素的升序排列。
sort(people.begin(), people.end(), compare);调用sort()便按照以上规则给队列排序。
result.insert(result.begin() + people[i].second, people[i]);此语句将序对插入到其第二个元素对应的下标位置。
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