2014-05-06 13:23
原题:
Finding a pair of elements from two sorted lists(or array) for which the sum of the elements is a certain value. Anyway solution that can do better than O(a.length + b.length)?
题目:给定两个有序的数组,如何从两数组中各选出一个元素使得两元素加起来等于某个目标值。
解法:又是这个“Guy”出的题目,此人想要追求优于O(n + m)的算法。这人代码水平不高,我想他对于算法复杂度的上下界也不知道怎么估计吧。我个人认为不太可能更优化了,因为你找的不是一个元素,而是一对。我的解法,是使用两个iterator,一个在A数组头部,一个在B数组尾部。通过A数组后移,B数组前移来调整相加的结果。这样的算法,复杂度就是O(n + m)的。
代码:
1 // http://www.careercup.com/question?id=6271724635029504 2 #include <cstdio> 3 #include <vector> 4 using namespace std; 5 6 bool twoSortedArraySum(vector<int> &a, vector<int> &b, int target, int &ia, int &ib) 7 { 8 int i, j; 9 int na, nb; 10 11 na = (int)a.size(); 12 nb = (int)b.size(); 13 14 i = 0; 15 j = nb - 1; 16 17 int sum; 18 while (i <= na - 1 && j >= 0) { 19 sum = a[i] + b[j]; 20 if (sum > target) { 21 --j; 22 } else if (sum < target) { 23 ++i; 24 } else { 25 ia = i; 26 ib = j; 27 return true; 28 } 29 } 30 return false; 31 } 32 33 int main() 34 { 35 vector<int> a, b; 36 int na, nb; 37 int i; 38 int ia, ib; 39 int target; 40 41 while (scanf("%d%d", &na, &nb) == 2 && (na > 0 && nb > 0)) { 42 a.resize(na); 43 b.resize(nb); 44 for (i = 0; i < na; ++i) { 45 scanf("%d", &a[i]); 46 } 47 for (i = 0; i < nb; ++i) { 48 scanf("%d", &b[i]); 49 } 50 while (scanf("%d", &target) == 1) { 51 ia = ib = -1; 52 if (twoSortedArraySum(a, b, target, ia, ib)) { 53 printf("%d + %d = %d ", a[ia], b[ib], target); 54 } else { 55 printf("Not found. "); 56 } 57 } 58 } 59 60 return 0; 61 }