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]]
这道题给了我们一个队列，队列中的每个元素是一个 pair，分别为身高和前面身高不低于当前身高的人的个数，让我们重新排列队列，使得每个 pair 的第二个参数都满足题意。首先来看一种超级简洁的方法，给队列先排个序，按照身高高的排前面，如果身高相同，则第二个数小的排前面。然后新建一个空的数组，遍历之前排好序的数组，然后根据每个元素的第二个数字，将其插入到 res 数组中对应的位置，参见代码如下：
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;
}
};
上面那种方法是简洁，但是用到了额外空间，我们来看一种不使用额外空间的解法，这种方法没有使用 vector 自带的 insert 或者 erase 函数，而是通过一个变量 cnt 和k的关系来将元素向前移动到正确位置，移动到方法是通过每次跟前面的元素交换位置，使用题目中给的例子来演示过程：
[[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]]
交换顺序：
[[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;
}
};
下面这种解法跟解法一很相似，只不过没有使用额外空间，而是直接把位置不对的元素从原数组中删除，直接加入到正确的位置上，参见代码如下：
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
类似题目：
Count of Smaller Numbers After Self
参考资料：
https://leetcode.com/problems/queue-reconstruction-by-height/
https://leetcode.com/problems/queue-reconstruction-by-height/discuss/89348/6-lines-Concise-C%2B%2B
LeetCode All in One 题目讲解汇总(持续更新中…)
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