剑指Offer——最小的K个数

 1 class Solution {
 2 public:
 3     vector<int> GetLeastNumbers_Solution(vector<int> input, int k) {
 4         int iSize = input.size();
 5         if(k == iSize) return input;
 6         vector<int> leastNumbers;
 7         if(k <= 0 || k > iSize) return leastNumbers;
 8         leastNumbers.push_back(0);  // 使下标为0的位置为0，不使用该位置，便于访问后面的元素
 9         for(int i = 0; i < k; i++) {    // 初始化一个k大小的堆
10             leastNumbers.push_back(input[i]);
11         }
12         for(int i = k / 2; i > 0; --i) {
13             HeapAdjust(leastNumbers, i, k); // 调整为大根堆
14         }
15         for(int i = k; i < iSize; i++) {
16             if(input[i] < leastNumbers[1]) {
17                 leastNumbers[1] = input[i];
18                 HeapAdjust(leastNumbers, 1, k);  // 调整为大根堆
19             }
20         }
21         leastNumbers.erase(leastNumbers.begin());   // 移除第0位置的0
22         return leastNumbers;
23     }
24 
25     void HeapAdjust(vector<int> &input, int s, int m) { // 调整堆
26         int rc = input[s];
27         for(int j = (s << 1); j <= m; j <<= 1) {
28             if(j < m && input[j] < input[j + 1]) ++j;
29             if(rc > input[j]) break;
30             input[s] = input[j];
31             s = j;
32         }
33         input[s] = rc;
34     }
35 };