406. Queue Reconstruction by Height

Description

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]]

简要题解：

这道题总的解题方向是采用贪心算法。具体做法是不断循坏people向量的各个元素直到该向量为空。每次从people中取出一个，在result中找到一个刚好可以放的位置放入。如果找不到可以放的位置放就跳过这个元素，等待下一轮循坏再继续找位置。

但是这样做会有点问题，我们必须保证后放入result的元素不会影响到先放入result的元素。因此，需要对people中的每个人按身高大小由高到底排序。（身高较低的不管怎么放都不会影响到身高较高的人）。还要注意到，对于两个身高相同的情况，k值小的应该放在前面。因为，身高相同的两个人，在result中k值小的一定是放在k值较大的那个人的前面。也就是说，在身高相同的情况下，k值较小的可能会影响k值较大的那个人。排完序后，就采用前面说的那种贪心算法来构造result就可得到正确的结果。

代码：

class Solution {public:    static vector<pair<int, int> > reconstructQueue(vector<pair<int, int> >& people) {        vector<pair<int, int> > result;        for (int i = 0; i < people.size(); 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;                }        while (!people.empty()) {            vector<pair<int, int> >::iterator iter1, iter2;            for (iter1 = people.begin(); iter1 != people.end(); iter1++) {                int count = 0;                int flag = false;                for (iter2 = result.begin(); iter2 != result.end(); iter2++) {                    if (count == iter1->second + 1) {                        iter2--;                        result.insert(iter2, *iter1);                        people.erase(iter1);                        flag = true;                        break;                    }                    if (iter2->first >= iter1->first)                        count++;                }                if (!flag && count == iter1->second + 1) {                    pair<int, int> temp = result[result.size()-1];                    result[result.size()-1] = *iter1;                    result.push_back(temp);                    people.erase(iter1);                    flag = true;                }                if (!flag && count == iter1->second) {                    result.push_back(*iter1);                    people.erase(iter1);                    flag = true;                }                if (flag)                    break;            }        }        return result;    }};