Careercup - Google面试题 - 4699414551592960

2014-05-06 13:34

题目链接

原题：

we have a random list of people. each person knows his own height and the number of tall people in front of him. write a code to make the equivalent queue. 
for example : 
input: <"Height","NumberOfTall","Name">, 
<6,2,"A">,<1,4,"B">,<11,0,"C">,<5,1,"D">,<10,0,"E">,<4,0,"F"> 
output: "F","E","D","C","B","A" --> end of queue

题目：有一队人正在排队，告诉你每个人的姓名、身高(目前默认都不相同)，以及他/她所看到的前面比他高的人。请你计算出这个队伍里各个人的顺序。

解法：从最矮的人入手。对于最矮的人，他所看到的所有人都比他高，所以如果他看到有K个人比他高，那么他一定排在K + 1名。我的代码是当时琢磨了很久，采用树状数组写出来的。此处的树状数组提供快速修改区间、查询单个元素的能力，能做到对数时间。现在居然已经忘了当时的具体思路。总体思想是先按身高升序排序，然后逐个确定每个人在队列里的位置。关键的一点：每次都只能确定当前最矮的人排在哪儿。树状数组中维护的值c[i]的意义，是记录当前位置的前面有多少个比它高。其中还有个方法用到二分搜索，所以整体复杂度是比较奇怪的O(n * log(n) +  n * log^2(n)) = O(n * log^2(n))。

代码：

// http://www.careercup.com/question?id=4699414551592960
#include <algorithm>
#include <iostream>
#include <string>
#include <vector>
using namespace std;

struct Person {
    int height;
    int taller;
    string name;
    Person(int _height = 0, int _taller = 0, string _name = ""): 
        height(_height), taller(_taller), name(_name) {};
    bool operator < (const Person &other) {
        return this->height < other.height;
    };
    
    friend ostream& operator << (ostream &cout, const Person &p) {
        cout << '<'<< p.name << p.height << '>';
        return cout;
    };
};

template <class T>
class BinaryIndexedTree {
public:
    BinaryIndexedTree(int _n = 0): n(_n), v(_n + 1) {};
    
    int size() {
        return n;
    };
    
    void addAll(int x, const T &val) {
        while (x >= 1 && x <= n) {
            v[x] += val;
            x -= lowBit(x);
        }
    };
    
    void addInterval(int x, int y, const T &val) {
        addAll(x - 1, -val);
        addAll(y, val);
    };
    
    void clear() {
        v.resize(1);
        n = 0;
    };
    
    void resize(int new_n) {
        v.resize(new_n + 1);
        n = new_n;
    }
    
    T sum(int x) {
        T res = 0;
        while (x >= 1 && x <= n) {
            res += v[x];
            x += lowBit(x);
        }
        
        return res;
    };
    
    int lowerBound(const T &val) {
        int ll, mm, rr;
        
        if (n == 0) {
            return 0;
        }

        T res;
        if (val <= (res = sum(1))) {
            return 1;
        }
        if (val > (res = sum(n))) {
            return n + 1;
        }
        
        ll = 1;
        rr = n;
        while (rr - ll > 1) {
            mm = (ll + rr) / 2;
            res = sum(mm);
            if (val > res) {
                ll = mm;
            } else {
                rr = mm;
            }
        }
        
        return rr;
    }
private:
    vector<T> v;
    int n;
    
    int lowBit(int x) {
        return x & -x;
    };
};

int main()
{
    vector<Person> people;
    vector<int> queue;
    BinaryIndexedTree<int> bit;
    int i, j;
    int n;
    
    while (cin >> n && n > 0) {
        people.resize(n);
        for (i = 0; i < n; ++i) {
            cin >> people[i].height >> people[i].taller >> people[i].name;
        }
        sort(people.begin(), people.end());
        bit.resize(n);
        queue.resize(n);
        for (i = 1; i <= n; ++i) {
            bit.addInterval(i, n, 1);
        }
        for (i = 1; i <= n; ++i) {
            j = bit.lowerBound(people[i - 1].taller + 1);
            queue[j - 1] = i;
            bit.addInterval(j, n, -1);
        }
        for (i = 0; i < n; ++i) {
            cout << people[queue[i] - 1];
        }
        cout << endl;
        
        people.clear();
        queue.clear();
        bit.clear();
    }
    
    return 0;
}