算法——Week13

406. 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]]

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解题思路
我先将vector中所有元素排序，h小的在前，h相同，k较大的在前（因为题目中k表示的是排在自己之前的大于等于的元素的个数，我在这里认为，k较大的是“小于”k较小的）。如题目给出的例子，排序后的结果如下：
[[4,4],[5,2],[5,0],[6,1],[7,1],[7,0]]
设置了一个标记位，标记这个元素是否放入结果中。
从排好序的第一个元素开始遍历，选择合适的位置。对于每个people[i]，因为之前的元素都已经放入，而且之前的元素都比它小，所以将其放入结果时，只需考虑流出k的空位置放比它大的元素。即选择第k个为空的位置。

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代码如下：

class Solution {
public:
    vector<pair<int, int>> reconstructQueue(vector<pair<int, int>>& people) {
        if(people.size() == 0) {
            return people;
        }
        for(int i = 0; i < people.size() - 1; i++) {
            for(int j = i + 1; j < people.size(); j++) {
                if(people[i].first > people[j].first || (people[i].first == people[j].first && people[i].second < people[j].second)) {
                pair<int, int> temp = people[i];
                people[i] = people[j];
                people[j] = temp;
                }   
            }
            
        }
        vector<int> tag(people.size());
        vector<pair<int, int>> res(people.size());
        for(int i = 0; i < people.size(); i++) {
            int j = 0, k = 0;
            for(j = 0, k = 0; j < people.size(); j++) {
                if(k != people[i].second && tag[j] != 0) {
                    continue;
                }
                if(k != people[i].second && tag[j] == 0) {
                    k++;
                    continue;
                }
                if(k == people[i].second && tag[j] == 0) {
                    res[j] = people[i];
                    tag[j] = 1;
                    break;
                } 
            }
        }
        return res;
    }
};