[leetcode]406. Queue Reconstruction by Height

greedy

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.

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

首先根据身高从高到低进行排序，如果身高相同，则根据k值从小到大进行排序。由于身高降序排序，则k值就是这个pair插入结果时所在的位置：res.insert(res.begin()+people[i].second, people[i])

排序后：

[[7,0], [7,1], [6,1], [5,0], [5,2], [4,4]]

1）将[7,0]插入结果

由于k = 0， 则插入结果中的第一个位置：[7,0]

2）将[7,1]插入结果

由于k = 1， 则插入结果中的第二个位置：[7,0]，[7,1]

3）将[6,1]插入结果

由于k = 1， 则插入结果中的第二个位置：[7,0]，[6,1]，[7,1]

4）将[5,0]插入结果

由于k = 0， 则插入结果中的第二个位置：[5, 0], [7,0]，[6,1]，[7,1]

代码：

class Solution {
public:
vector<pair<int, int>> reconstructQueue(vector<pair<int, int>>& people) {
sort(people.begin(), people.end(), cmp);
vector<pair<int, int>> res;
for (int i = 0; i < people.size(); i++) {
res.insert(res.begin()+people[i].second, people[i]);
}
return res;
}
private:
static int cmp(pair<int, int> a, pair<int, int> b) {
if (a.first > b.first || (a.first == b.first && a.second <= b.second)) return true;
else return<
8c01
/span> false;
}
};