[leetcode-406-Queue Reconstruction by Height]

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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 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：

参考的别人的，下边有链接，

1. Pick out tallest group of people and sort them in a subarray (S). Since there‘s no other groups of people taller than them, therefore each guy‘s index will be just as same as his k value.
2. For 2nd tallest group (and the rest), insert each one of them into (S) by k value. So on and so forth.

E.g.
input: [[7,0], [4,4], [7,1], [5,0], [6,1], [5,2]]
subarray after step 1: [[7,0], [7,1]]
subarray after step 2: [[7,0], [6,1], [7,1]]
...

It‘s not the most concise code, but I think it well explained the concept.

vector<pair<int, int>> reconstructQueue(vector<pair<int, int>>& people) {
    auto comp = [](const pair<int, int>& p1, const pair<int, int>& p2)
                    { return p1.first > p2.first || (p1.first == p2.first && p1.second < p2.second); };
    sort(people.begin(), people.end(), comp);
    vector<pair<int, int>> res;
    for (auto& p : people) 
        res.insert(res.begin() + p.second, p);
    return res;
}

参考：

https://discuss.leetcode.com/topic/60394/easy-concept-with-python-c-java-solution

https://discuss.leetcode.com/topic/60470/6-lines-concise-c

[leetcode-406-Queue Reconstruction by Height]

标签：pytho   front   tar   xpl   http   therefore   ==   cto   step   

原文地址：http://www.cnblogs.com/hellowooorld/p/6896768.html