1 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,。
2 思路和方法,C++核心代码
2.1 sort()函数,vector<int> push_back(); nlog(n)
1 class Solution { 2 public: 3 vector<int> GetLeastNumbers_Solution(vector<int> input, int k) { 4 sort(input.begin(),input.end()); 5 vector<int> result; 6 if(k>input.size()) 7 return result; 8 for(int i = 0;i<k;i++){ 9 result.push_back(input[i]); 10 } 11 return result; 12 } 13 };
2.2 Partition分治O(n)
1 class Solution { 2 public: 3 vector<int> GetLeastNumbers_Solution(vector<int> input, int k) { 4 // 必须检查k的大小否则出错!!! 5 if (input.size() == 0 || k > input.size() || k<=0) 6 return vector<int>(); 7 8 int start = 0; 9 int end = input.size() -1; 10 int index = Partition(input,start,end); 11 while(index != k-1){ 12 if(index < k-1){ 13 start = index + 1; // k在index右边 14 index = Partition(input,start,end); 15 } 16 else{ 17 end = index -1; // k在index左边 18 index = Partition(input,start,end); 19 } 20 } 21 vector<int> ret; 22 for (int i = 0; i < k; ++i) { 23 ret.push_back(input[i]); 24 } 25 return ret; 26 } 27 28 int Partition(vector<int> &data, int start, int end){ 29 if(start > end) 30 return start; 31 32 int key = data[start]; // 取第一个值为参考值 33 while(start< end){ 34 while(data[end] >= key && start < end) end--; 35 data[start] = data[end]; 36 while(data[start] <= key && start<end) start++; 37 data[end] = data[start]; 38 } 39 data[start] = key; 40 return start; 41 } 42 };
参考资料
https://blog.csdn.net/zjwreal/article/details/88608506