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【ACWing】53. 最小的k个数
阅读量:241 次
发布时间:2019-02-28

本文共 1174 字,大约阅读时间需要 3 分钟。

题目地址:

输入 n n n个整数,找出其中最小的 k k k个数。

注意:输出数组内元素请按从小到大顺序排序。

数据范围:

1 ≤ k ≤ n ≤ 1000 1≤k≤n≤1000 1kn1000

可以用quick select算法将前 k k k小的数找出来,然后对前 k k k个数排序即可。代码如下:

#include 
#include
#include
using namespace std;class Solution { public: void quick_select(vector
& v, int l, int r, int idx) { if (l >= r) return; int x = v[l + (r - l >> 1)], i = l, j = r; while (i <= j) { while (v[i] < x) i++; while (v[j] > x) j--; if (i <= j) { swap(v[i], v[j]); i++; j--; } } if (idx <= j) quick_select(v, l, j, idx); else if (idx >= i) quick_select(v, i, r, idx); else return; } vector
getLeastNumbers_Solution(vector
input, int k) { quick_select(input, 0, input.size() - 1, k - 1); sort(input.begin(), input.begin() + k); vector
res; for (int i = 0; i < k; i++) res.push_back(input[i]); return res; }};

平均时间复杂度 O ( n + k log ⁡ k ) O(n+k\log k) O(n+klogk),平均空间 O ( log ⁡ n ) O(\log n) O(logn)

转载地址:http://xqjs.baihongyu.com/

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