《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：

 1 // 11.1 Given two sorted array A and B, suppose A is large enough to hold them both. Merge B into A.
 2 #include <algorithm>
 3 #include <cstdio>
 4 using namespace std;
 5 
 6 void mergeBIntoA(int a[], int b[], int na, int nb)
 7 {
 8     if (na <= 0 || nb <= 0) {
 9         // invalid parameter
10         return;
11     }
12     
13     int *pmin, *pmax;
14     pmin = min(a, b);
15     pmax = max(a + na + nb, b + nb);
16     if (pmax - pmin < na + nb + nb) {
17         // the memories overlap
18         return;
19     }
20     
21     int i, j, k;
22     
23     i = na;
24     j = nb;
25     k = na + nb;
26     
27     while (i > 0 && j > 0) {
28         if (a[i - 1] > b[j - 1]) {
29             a[--k] = a[--i];
30         } else {
31             a[--k] = b[--j];
32         }
33     }
34     while (j > 0) {
35         a[--k] = b[--j];
36     }
37 }
38 
39 int main()
40 {
41     int *a, *b;
42     int na, nb;
43     int i;
44     
45     while (scanf("%d%d", &na, &nb) ==  2 && (na > 0 && nb > 0)) {
46         a = new int[na + nb];
47         b = new int[nb];
48         for (i = 0; i < na; ++i) {
49             scanf("%d", &a[i]);
50         }
51         for (i = 0; i < nb; ++i) {
52             scanf("%d", &b[i]);
53         }
54         mergeBIntoA(a, b, na, nb);
55         for (i = 0; i < na + nb; ++i) {
56             printf((i == 0 ? "%d" : " %d"), a[i]);
57         }
58         putchar('\n');
59         
60         delete[] a;
61         delete[] b;
62         a = nullptr;
63         b = nullptr;
64     }
65     
66     return 0;
67 }

 

 posted on 2014-03-21 20:48  zhuli19901106  阅读(277)  评论(0)  编辑  收藏  举报