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 = 0;
        n = (int)sticks.size();
        for (i = 0; i < n; ++i) {
            pq.push(sticks[i]);
        }
        
        for (i = 0; i < n - 1; ++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;
};