Careercup

2014-05-02 07:37

题目链接

原题：

// merge sorted arrays 'a' and 'b', each with 'length' elements, 
// in-place into 'b' to form a sorted result. assume that 'b' 
// has 2*length allocated space. 
// e.g. a = [1, 3, 5], b = [2, 4, 6] => b = [1, 2, 3, 4, 5, 6] 

//how to do it without rearanging the b array

题目：有两个排好序的数组a[]和b[]，把a有序归并到b中去，保证b的空间充足。如何就地完成这个算法？

解法：从后往前归并就可以不用额外空间了。

代码：

// http://www.careercup.com/question?id=5435439490007040
#include <iostream>
#include <vector>
using namespace std;

class Solution {
public:
    void mergeTwoArray (vector<int> &a, vector<int> &b) {
        // merge a[] into b[].
        int na = (int)a.size();
        int nb = (int)b.size();

        b.resize(na + nb);

        int i, j, k;
        
        i = na - 1;
        j = nb - 1;
        k = na + nb - 1;
        while (i >= 0 && j >= 0) {
            b[k--] = a[i] > b[j] ? a[i--] : b[j--];
        }
        while (i >= 0) {
            b[k--] = a[i--];
        }
    };
};

int main()
{
    vector<int> a, b;
    int na, nb;
    int i;
    Solution sol;
    
    while (cin >> na >> nb && (na > 0 && nb > 0)) {
        a.resize(na);
        b.resize(nb);
        for (i = 0; i < na; ++i) {
            cin >> a[i];
        }
        for (i = 0; i < nb; ++i) {
            cin >> b[i];
        }
        sol.mergeTwoArray(a, b);
        nb = (int)b.size();
        for (i = 0; i < nb; ++i) {
            i ? (cout << ' ', 1) : 1;
            cout << b[i];
        }
        cout << endl;
        
        a.clear();
        b.clear();
    }
    
    return 0;
}