《Cracking the Coding Interview》——第11章：排序和搜索——题目1

2014-03-21 20:35

题目：给定已升序排列的数组A和数组B，如果A有足够的额外空间容纳A和B，请讲B数组合入到A中。

解法：由后往前进行归并。

代码：

// 11.1 Given two sorted array A and B, suppose A is large enough to hold them both. Merge B into A.
#include <algorithm>
#include <cstdio>
using namespace std;

void mergeBIntoA(int a[], int b[], int na, int nb)
{
    if (na <= 0 || nb <= 0) {
        // invalid parameter
        return;
    }
    
    int *pmin, *pmax;
    pmin = min(a, b);
    pmax = max(a + na + nb, b + nb);
    if (pmax - pmin < na + nb + nb) {
        // the memories overlap
        return;
    }
    
    int i, j, k;
    
    i = na;
    j = nb;
    k = na + nb;
    
    while (i > 0 && j > 0) {
        if (a[i - 1] > b[j - 1]) {
            a[--k] = a[--i];
        } else {
            a[--k] = b[--j];
        }
    }
    while (j > 0) {
        a[--k] = b[--j];
    }
}

int main()
{
    int *a, *b;
    int na, nb;
    int i;
    
    while (scanf("%d%d", &na, &nb) ==  2 && (na > 0 && nb > 0)) {
        a = new int[na + nb];
        b = new int[nb];
        for (i = 0; i < na; ++i) {
            scanf("%d", &a[i]);
        }
        for (i = 0; i < nb; ++i) {
            scanf("%d", &b[i]);
        }
        mergeBIntoA(a, b, na, nb);
        for (i = 0; i < na + nb; ++i) {
            printf((i == 0 ? "%d" : " %d"), a[i]);
        }
        putchar('\n');
        
        delete[] a;
        delete[] b;
        a = nullptr;
        b = nullptr;
    }
    
    return 0;
}