《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 <=  || nb <= ) {
         // 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 >  && j > ) {
         if (a[i - ] > b[j - ]) {
             a[--k] = a[--i];
         } else {
             a[--k] = b[--j];
         }
     }
     while (j > ) {
         a[--k] = b[--j];
     }
 }
 
 int main()
 {
     int *a, *b;
     int na, nb;
     int i;
 
     while (scanf("%d%d", &na, &nb) ==   && (na >  && nb > )) {
         a = new int[na + nb];
         b = new int[nb];
         for (i = ; i < na; ++i) {
             scanf("%d", &a[i]);
         }
         for (i = ; i < nb; ++i) {
             scanf("%d", &b[i]);
         }
         mergeBIntoA(a, b, na, nb);
         for (i = ; i < na + nb; ++i) {
             printf((i ==  ? "%d" : " %d"), a[i]);
         }
         putchar('\n');
 
         delete[] a;
         delete[] b;
         a = nullptr;
         b = nullptr;
     }
 
     return ;
 }