我要寻找的解决方案:
Given a array and a number P , find two numbers in array whose product equals P.
寻找比O解决方案更好的(N * 2)。我没关系使用额外的空间或其他数据结构。任何帮助是AP preciated?
Looking for solution better than O(n*2) . I am okay with using extra space or other datastructure .Any help is appreciated ?
推荐答案您可以尝试滑窗的方法。首先排序所有的数字越来越多,然后用两个整数开始和结束指数目前对数字。初始化开始 0和结束到最后的位置。然后比较 v [开始] 和 v [结束] 与 P :
You can try a sliding window approach. First sort all the numbers increasingly, and then use two integers begin and end to index the current pair of numbers. Initialize begin to 0 and end to the last position. Then compare the product of v[begin] and v[end] with P:
- 如果是平等的,你找到了答案。
- 如果是较低的,你必须找到一个更大的产品,移动开始前进。
- 如果是更高的,你必须找到一个更小的产品,移动结束落后。
- If it is equal, you found the answer.
- If it is lower, you must find a bigger product, move begin forward.
- If it is higher, you must find a smaller product, move end backward.
下面是一个C ++ code这个想法实现。该解决方案是O,因为排序(N *的log(n)),如果你可以假设数据进行排序,那么你可以跳过一个O排序(N)解决方案。
Here is a C++ code with this idea implemented. This solution is O(n*log(n)) because of the sorting, if you can assume the data is sorted then you can skip the sorting for an O(n) solution.
pair<int, int> GetProductPair(vector<int>& v, int P) { sort(v.begin(), v.end()); int begin = 0, end = static_cast<int>(v.size()) - 1; while (begin < end) { const int prod = v[begin] * v[end]; if (prod == P) return make_pair(begin, end); if (prod < P) ++begin; else --end; } return make_pair(-1, -1); }更多推荐
给定一个数p,发现数组,其产品的两个元素= P
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