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]]
这个题怎么想出来的呢?是因为我们考虑如果先把个子高的排好序,那么在任何位置插入数据都不会对已经排好序的数组造成影响。而,与此同时,我们已经知道了个子高的排序,那么当新的数据到的时候,我们要确定它的位置也很简单,因为现在的所有数据都比他高,所以只要根据他的第二个数字确定他的位置即可。
先对已有的数组进行排序。按照高度降序排列,如果高度一样,按照k的值升序排列。这样比如一开始[7,0] [7,1] [7,2]就会排好,然后比如说后面有一个[6,1], 说明只有一个大于或等于它,又因为比6大的已经全部取出。所以把它放在位置1,这样就变成[7,0] [6,1] [7,1] [7,2].然后比如又有一个[5,0].就放在位置0,以此类推。
即对于案列。首先排序成:
[[7,0],[7,1],[6,1],[5,0],[5,2],[4,4]]
然后对于第二个数字进行插入对应位置:
[[5,0], [7,0], [5,2], [6,1], [4,4], [7,1]]
注:这道题思路不难想,难在第一次接触这种list队依据两个因素排序。
Code
class Solution:
def reconstructQueue(self, people):
"""
:type people: List[List[int]]
:rtype: List[List[int]]
"""
if not people or not people[0]:
return []
temp_people = sorted(people, key = lambda people:(-people[0], people[1]))
res = []
for man in temp_people:
res.insert(man[1], man)
return res
C++代码如下,需要注意自定义sort()函数的比较方法。
注意这种自定义sort函数的方式。
class Solution {
public:
vector<pair<int, int>> reconstructQueue(vector<pair<int, int>>& people) {
sort(people.begin(), people.end(), [](const pair<int, int> a, const pair<int, int> b) {
return (a.first > b.first) || (a.first == b.first && a.second < b.second);
});
vector<pair<int, int>> res;
for (auto p: people){
res.insert(res.begin()+p.second, p);
}
return res;
}
};
people.sort(key = lambda x : (-x[0], x[1]))
或 temp_people = sorted(people, key = lambda people:(-people[0], people[1]))
,很值得学习!sort(people.begin(), people.end(), [](const pair a, const pair b) {
return (a.first > b.first) || (a.first == b.first && a.second < b.second);
});
[1] https://blog.csdn.net/fuxuemingzhu/article/details/68486884