Careercup

2014-05-12 07:31

原题：

I have heard this question many times in microsoft interviews. Given two arrays find the intersection of those two arrays. Besides using hash table can we attain the same time complexity that is O(m+n) by using some other approach.

题目：给定两个数组，计算出他们的交集。要求线性时间完成。

解法1：出题者问能否在不用哈希的条件下，用线性时间完成。数组本身不一定是有序的，所以我没想出不用哈希的线性算法。一种可行的解法，是先排序两个数组，然后进行归并，取其交集。显然这种算法的复杂度主要来自排序。

代码：

 1 // http://www.careercup.com/question?id=24308662
 2 // nobody said the elements in both arrays are unique, why would bitset work?
 3 #include <algorithm>
 4 #include <iostream>
 5 #include <vector>
 6 using namespace std;
 7 
 8 class Solution {
 9 public:
10     void mergeIntersection(vector<int> &a1, vector<int> &a2, vector<int> &intersect) {
11         sort(a1.begin(), a1.end());
12         sort(a2.begin(), a2.end());
13         
14         int i, j;
15         int n1, n2;
16         
17         i = 0;
18         j = 0;
19         n1 = (int)a1.size();
20         n2 = (int)a2.size();
21         intersect.clear();
22         while (i < n1 && j < n2) {
23             if (a1[i] < a2[j]) {
24                 ++i;
25             } else if (a1[i] > a2[j]) {
26                 ++j;
27             } else {
28                 intersect.push_back(a1[i]);
29                 ++i;
30                 ++j;
31             }
32         }
33     };
34 };
35 
36 int main()
37 {
38     int n1, n2, n;
39     vector<int> a1, a2;
40     vector<int> intersect;
41     int i;
42     Solution sol;
43     
44     while (cin >> n1 >> n2 && (n1 > 0 && n2 > 0)) {
45         a1.resize(n1);
46         a2.resize(n2);
47         for (i = 0; i < n1; ++i) {
48             cin >> a1[i];
49         }
50         for (i = 0; i < n2; ++i) {
51             cin >> a2[i];
52         }
53         sol.mergeIntersection(a1, a2, intersect);
54         
55         cout << '{';
56         n = (int)intersect.size();
57         for (i = 0; i < n; ++i) {
58             i ? (cout << ' '), 1 : 1;
59             cout << intersect[i];
60         }
61         cout << '}' << endl;
62     }
63     
64     return 0;
65 }

解法2：用哈希来搞定，可以在线性时间内完成。统计两个数组中每个值出现的次数，取较少的次数作为交集。比如A[]中有3个“1”，B[]中有5个“1”，那么交集就有3个“1”。

代码：

 1 // http://www.careercup.com/question?id=24308662
 2 // nobody said the elements in both arrays are unique, why would bitset work?
 3 #include <algorithm>
 4 #include <iostream>
 5 #include <vector>
 6 #include <unordered_map>
 7 using namespace std;
 8 
 9 class Solution {
10 public:
11     void mergeIntersection(vector<int> &a1, vector<int> &a2, vector<int> &intersect) {
12         if (a1.size() > a2.size()) {
13             mergeIntersection(a2, a1, intersect);
14             return;
15         }
16         unordered_map<int, pair<int, int> > um;
17         int n1, n2;
18         int i;
19         
20         n1 = (int)a1.size();
21         n2 = (int)a2.size();
22         unordered_map<int, pair<int, int> >::iterator it;
23         
24         for (i = 0; i < n1; ++i) {
25             it = um.find(a1[i]);
26             if (it == um.end()) {
27                 um[a1[i]] = make_pair(1, 0);
28             } else {
29                 ++it->second.first;
30             }
31         }
32         
33         for (i = 0; i < n2; ++i) {
34             it = um.find(a2[i]);
35             if (it != um.end()) {
36                 ++it->second.second;
37             }
38         }
39         
40         intersect.clear();
41         for (it = um.begin(); it != um.end(); ++it) {
42             n1 = min(it->second.first, it->second.second);
43             for (i = 0; i < n1; ++i) {
44                 intersect.push_back(it->first);
45             }
46         }
47         
48         um.clear();
49     };
50 };
51 
52 int main()
53 {
54     int n1, n2, n;
55     vector<int> a1, a2;
56     vector<int> intersect;
57     int i;
58     Solution sol;
59     
60     while (cin >> n1 >> n2 && (n1 > 0 && n2 > 0)) {
61         a1.resize(n1);
62         a2.resize(n2);
63         for (i = 0; i < n1; ++i) {
64             cin >> a1[i];
65         }
66         for (i = 0; i < n2; ++i) {
67             cin >> a2[i];
68         }
69         sol.mergeIntersection(a1, a2, intersect);
70         
71         cout << '{';
72         n = (int)intersect.size();
73         for (i = 0; i < n; ++i) {
74             i ? (cout << ' '), 1 : 1;
75             cout << intersect[i];
76         }
77         cout << '}' << endl;
78     }
79     
80     return 0;
81 }