面试题30:最小的k个数
题目描述
输入n个整数,找出其中最小的k个数。例如输入4、5、1、6、2、7、3、8这8个数字,则最小的4个数字是1、2、3、4。
题目分析
剑指Offer(纪念版)P167
代码实现
O(n)的算法,只有当我们可以修改输入的数组时可用
void GetLeastNumbers_Solution1(int* input, int n, int* output, int k) { if(input == NULL || output == NULL || k > n || n <= 0 || k <= 0) return; int start = 0; int end = n - 1; int index = Partition(input, n, start, end); while(index != k - 1) { if(index > k - 1) { end = index - 1; index = Partition(input, n, start, end); } else { start = index + 1; index = Partition(input, n, start, end); } } for(int i = 0; i < k; ++i) output[i] = input[i]; } int Partition(int data[], int length, int start, int end) { if(data == NULL || length <= 0 || start < 0 || end >= length) throw new std::exception("Invalid Parameters"); int index = RandomInRange(start, end); Swap(&data[index], &data[end]); int small = start - 1; for(index = start; index < end; ++ index) { if(data[index] < data[end]) { ++ small; if(small != index) Swap(&data[index], &data[small]); } } ++ small; Swap(&data[small], &data[end]); return small; } // Random Partition int RandomInRange(int min, int max) { int random = rand() % (max - min + 1) + min; return random; } void Swap(int* num1, int* num2) { int temp = *num1; *num1 = *num2; *num2 = temp; }
O(nlogk)的算法,特别适合处理海量数据
下面的代码也可以使用priority_queue来实现
typedef multiset<int, greater<int> > intSet; typedef multiset<int, greater<int> >::iterator setIterator; void GetLeastNumbers_Solution2(const vector<int>& data, intSet& leastNumbers, int k) { leastNumbers.clear(); if(k < 1 || data.size() < k) return; vector<int>::const_iterator iter = data.begin(); for(; iter != data.end(); ++ iter) { if((leastNumbers.size()) < k) leastNumbers.insert(*iter); else { setIterator iterGreatest = leastNumbers.begin(); if(*iter < *(leastNumbers.begin())) { leastNumbers.erase(iterGreatest); leastNumbers.insert(*iter); } } } }