标签:style class blog code java http
最简单的方法是将n个元素排序,取出最小的k个元素。这个算法的时间复杂度为 O(nlgn)。
然而在输入的n个元素互异的情况下,利用最大堆,我们可以获得时间复杂度为 O(nlgk)的算法。
1 #include <stdio.h> 2 3 #define N 128 4 5 int heap[N], max_size, cur_pos = 1; 6 7 void adjust(int i) { // bottom up 8 int x = heap[i]; 9 int p = i / 2; 10 while (p != 0) { 11 if (x > heap[p]) { // max heap 12 heap[i] = heap[p]; 13 i = p; 14 p = i / 2; 15 } else break; 16 } 17 heap[i] = x; 18 } 19 20 void adjust2(int x) { // top down 21 int i = 1, lim = max_size / 2; 22 if (x > heap[1]) return; 23 while (i <= lim) { 24 int L = i * 2, R = i * 2 + 1; 25 int tmp = heap[L], j = L; 26 if (R <= max_size && heap[R] > tmp) { 27 tmp = heap[R], j = R; 28 } 29 if (tmp > x) { 30 heap[i] = tmp, i = j; 31 } else break; 32 } 33 heap[i] = x; 34 } 35 36 void insert(int x) { 37 if (cur_pos <= max_size) { 38 heap[cur_pos] = x; 39 adjust(cur_pos); 40 cur_pos++; 41 } else { 42 adjust2(x); 43 } 44 } 45 46 void print() { 47 int i; 48 for (i = 1; i < cur_pos; i++) { 49 printf("%d ", heap[i]); 50 } 51 printf("\n"); 52 } 53 54 55 int main() { 56 int data[10] = {1, 4, 2, 8, 5, 7, 9, 3, 0, 6}; 57 int n = 10, k = 5; 58 int i; 59 60 max_size = k; 61 for (i = 0; i < n; i++) { 62 insert(data[i]); // O(lgk) 63 print(); 64 } 65 66 return 0; 67 }
3.20 采购订单自动生成销售订单,布布扣,bubuko.com
标签:style class blog code java http
原文地址:http://www.cnblogs.com/st-sun/p/3772837.html