Careercup - Google面试题 - 4557716425015296

2014-05-03 21:57

题目链接

原题：

Many sticks with length, every time combine two, the cost is the sum of two sticks' length. Finally, it will become a stick, what's the minimum cost?

题目：有很多根长度不同的棍子，每次选择其中两根拼起来，每次拼接的代价为两木棍长度之和。问把所有棍子拼成一根最少需要花多大代价。

解法：描述这么麻烦，还不如直接问哈夫曼编码呢。根据贪婪原则，每次选取长度最短的两个木棍即可。用小顶堆可以持续进行这个过程，直到只剩一根木棍为止。由于单个堆操作是对数级的，所以算法总体复杂度是O(n * log(n))，空间复杂度为O(n)。仿函数greater和less，对应于小顶堆和大顶堆。起初我经常搞反，时间长了就记住了。

代码：

 1 // http://www.careercup.com/question?id=4557716425015296

 2 #include <queue>

 3 #include <vector>

 4 using namespace std;

 5 

 6 template <class T>

 7 struct greater {

 8     bool operator () (const T &x, const T &y) {

 9         return x > y;

10     };

11 };

12 

13 class Solution {

14 public:

15     int minimalLengthSum(vector<int> &sticks) {

16         int i, n;

17         int sum;

18         int num1, num2;

19         

20         sum = 0;

21         n = (int)sticks.size();

22         for (i = 0; i < n; ++i) {

23             pq.push(sticks[i]);

24         }

25         

26         for (i = 0; i < n - 1; ++i) {

27             num1 = pq.top();

28             pq.pop();

29             num2 = pq.top();

30             pq.pop();

31             sum += num1 + num2;

32             pq.push(num1 + num2);

33         }

34         

35         while (!pq.empty()) {

36             pq.pop();

37         }

38         

39         return sum;

40     };

41 private:

42     // min heap

43     priority_queue<int, vector<int>, greater<int> > pq;

44 };