[LeetCode] 406. Queue Reconstruction by Height 根據高度重建隊列

 

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]]

 

這道題給了我們一個隊列，隊列中的每個元素是一個 pair，分別為身高和前面身高不低於當前身高的人的個數，讓我們重新排列隊列，使得每個 pair 的第二個參數都滿足題意。首先來看一種超級簡潔的方法，給隊列先排個序，按照身高高的排前面，如果身高相同，則第二個數小的排前面。然后新建一個空的數組，遍歷之前排好序的數組，然后根據每個元素的第二個數字，將其插入到 res 數組中對應的位置，參見代碼如下：

 

解法一：

class Solution {
public:
    vector<vector<int>> reconstructQueue(vector<vector<int>>& people) {
        sort(people.begin(), people.end(), [](vector<int>& a, vector<int>& b) {
            return a[0] > b[0] || (a[0] == b[0] && a[1] < b[1]);
        });
        vector<vector<int>> res;
        for (auto a : people) {
            res.insert(res.begin() + a[1], a);
        }
        return res;
    }
};

 

上面那種方法是簡潔，但是用到了額外空間，我們來看一種不使用額外空間的解法，這種方法沒有使用 vector 自帶的 insert 或者 erase 函數，而是通過一個變量 cnt 和k的關系來將元素向前移動到正確位置，移動到方法是通過每次跟前面的元素交換位置，使用題目中給的例子來演示過程：

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

排序后：

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

交換順序：

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

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

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

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

 

解法二：

class Solution {
public:
    vector<vector<int>> reconstructQueue(vector<vector<int>>& people) {
        sort(people.begin(), people.end(), [](vector<int>& a, vector<int>& b) {
            return a[0] > b[0] || (a[0] == b[0] && a[1] < b[1]);
        });
        for (int i = 1; i < people.size(); ++i) {
            int cnt = 0;
            for (int j = 0; j < i; ++j) {
                if (cnt == people[i][1]) {
                    auto t = people[i];
                    for (int k = i - 1; k >= j; --k) {
                        people[k + 1] = people[k];
                    }
                    people[j] = t;
                    break;
                }
                ++cnt;
            }
        }
        return people;
    }
};

 

下面這種解法跟解法一很相似，只不過沒有使用額外空間，而是直接把位置不對的元素從原數組中刪除，直接加入到正確的位置上，參見代碼如下：

 

解法三：

class Solution {
public:
    vector<vector<int>> reconstructQueue(vector<vector<int>>& people) {
        sort(people.begin(), people.end(), [](vector<int>& a, vector<int>& b) {
            return a[0] > b[0] || (a[0] == b[0] && a[1] < b[1]);
        });
        for (int i = 0; i < people.size(); i++) {
            auto p = people[i];
            if (p[1] != i) {
                people.erase(people.begin() + i);
                people.insert(people.begin() + p[1], p);
            }
        }
        return people;
    }
};

 

Github 同步地址：

https://github.com/grandyang/leetcode/issues/406

 

類似題目：

Count of Smaller Numbers After Self

 

參考資料：

https://leetcode.com/problems/queue-reconstruction-by-height/

https://leetcode.com/problems/queue-reconstruction-by-height/discuss/89348/6-lines-Concise-C%2B%2B

https://leetcode.com/problems/queue-reconstruction-by-height/discuss/89456/short-java-solution-without-using-extra-space

https://leetcode.com/problems/queue-reconstruction-by-height/discuss/89345/Easy-concept-with-PythonC%2B%2BJava-Solution

 

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