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：

 // http://www.careercup.com/question?id=4557716425015296
 #include <queue>
 #include <vector>
 using namespace std;

 template <class T>
 struct greater {
     bool operator () (const T &x, const T &y) {
         return x > y;
     };
 };

 class Solution {
 public:
     int minimalLengthSum(vector<int> &sticks) {
         int i, n;
         int sum;
         int num1, num2;

         sum = ;
         n = (int)sticks.size();
         for (i = ; i < n; ++i) {
             pq.push(sticks[i]);
         }

         for (i = ; i < n - ; ++i) {
             num1 = pq.top();
             pq.pop();
             num2 = pq.top();
             pq.pop();
             sum += num1 + num2;
             pq.push(num1 + num2);
         }

         while (!pq.empty()) {
             pq.pop();
         }

         return sum;
     };
 private:
     // min heap
     priority_queue<int, vector<int>, greater<int> > pq;
 };