《Cracking the Coding Interview》——第17章：普通题——题目12

2014-04-29 00:04

题目：给定一个整数数组，找出所有加起来为指定和的数对。

解法1：可以用哈希表保存数组元素，做到O(n)时间的算法。

代码：

 1 // 17.12 Given an array of integers and target value, find all pairs in the array that sum up to the target.

 2 // Use hash to achieve O(n) time complexity. Duplicates pairs are skipped.

 3 #include <cstdio>

 4 #include <unordered_map>

 5 #include <vector>

 6 using namespace std;

 7 

 8 int main()

 9 {

10     vector<int> v;

11     unordered_map<int, int> um;

12     int n, i;

13     int x, y;

14     int target;

15     

16     while (scanf("%d", &n) == 1 && n > 0) {

17         scanf("%d", &target);

18         

19         v.resize(n);

20         for (i = 0; i < n; ++i) {

21             scanf("%d", &v[i]);

22         }

23         

24         // duplicate pairs are skipped

25         for (i = 0; i < n; ++i) {

26             um[v[i]] = um[v[i]] + 1;

27         }

28         

29         unordered_map<int, int>::iterator it, it2;

30         for (it = um.begin(); it != um.end(); ++it) {

31             x = it->first;

32             y = target - x;

33             if (x > y) {

34                 continue;

35             }

36             

37             --it->second;

38             if ((it2 = um.find(y)) != um.end() && it2->second > 0) {

39                 printf("(%d, %d)\n", x, y);

40             }

41             ++it->second;

42         }

43         

44         v.clear();

45         um.clear();

46     }

47     

48     return 0;

49 }

解法2：先将数组排序，然后用两个iterator向中间靠拢进行扫描。总体时间是O(n * log(n))。

代码：

 1 // 17.12 Given an array of integers and target value, find all pairs in the array that sum up to the target.

 2 // O(n * log(n) + n) solution.

 3 #include <algorithm>

 4 #include <cstdio>

 5 #include <vector>

 6 using namespace std;

 7 

 8 int main()

 9 {

10     vector<int> v;

11     int n, i;

12     int ll, rr;

13     int target;

14     

15     while (scanf("%d", &n) == 1 && n > 0) {

16         scanf("%d", &target);

17         

18         v.resize(n);

19         for (i = 0; i < n; ++i) {

20             scanf("%d", &v[i]);

21         }

22         sort(v.begin(), v.end());

23         ll = 0;

24         rr = n - 1;

25         

26         int sum;

27         while (ll < rr) {

28             sum = v[ll] + v[rr];

29             if (sum < target) {

30                 while (ll + 1 < rr && v[ll] == v[ll + 1]) {

31                     ++ll;

32                 }

33                 ++ll;

34             } else if (sum > target) {

35                 while (rr - 1 > ll && v[rr] == v[rr - 1]) {

36                     --rr;

37                 }

38                 --rr;

39             } else {

40                 printf("(%d, %d)\n", v[ll], v[rr]);

41                 while (ll + 1 < rr && v[ll] == v[ll + 1]) {

42                     ++ll;

43                 }

44                 ++ll;

45             }

46         }

47         

48         v.clear();

49     }

50     

51     return 0;

52 }