剑指offer第二十九题最小的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

题风一转，简单易行，直接上代码吧。可以库函数sort，快排思想和堆排序

class Solution {
public:
    vector<int> GetLeastNumbers_Solution(vector<int> input, int k) {
        int size=input.size();
        vector<int> ret;
        if(input.empty()||k>input.size()) return ret;
        sort(input.begin(),input.end());
        for(int i=0;i<k;i++){
            ret.push_back(input[i]);
        }
        return ret;
    }
};

快排

class Solution {
public:
    void swap(int &fir,int &sec)
    {
        int temp = fir;
        fir = sec;
        sec = temp;
    }
     
    int getPartition(vector<int> &input,int start,int end)
    {
        if(input.empty() || start>end) return -1;
        int temp = input[end];
        int j = start - 1;
        for(int i=start;i<end;++i)
        {
            if(input[i]<=temp)
            {
                ++j;
                if(i!=j) swap(input[i],input[j]);                   
            }
        }
        swap(input[j+1],input[end]);
        return (j+1);
    }
         
    vector<int> GetLeastNumbers_Solution(vector<int> input, int k)
    {
        vector<int> result;       
        if(input.empty() || k>input.size() || k<=0) return result;
         
        int start = 0;
        int end = input.size()-1;
        int index = getPartition(input,start,end);
         
        while(index != (k-1))
        {
            if(index > (k-1))
            {
                end = index - 1;
                index = getPartition(input,start,end);
            }
            else
            {
                start = index + 1;
                index = getPartition(input,start,end);
            }
        }
         
        for(int i=0;i<k;++i)
        {
            result.push_back(input[i]);
        }
         
        return result;
    }
};

堆排序

class Solution {
public:
    vector<int> GetLeastNumbers_Solution(vector<int> input, int k)
    {
        vector<int> result;
        int len = input.size();
        if(input.empty() || k<=0 || len < k) return result;
         
        multiset<int, greater<int> > leastNumbers; // 从大到小排序
        multiset<int, greater<int> >::iterator iterGreater; // 定义迭代器
         
        vector<int>::iterator iter = input.begin();
        for(; iter != input.end(); ++iter)
        {
            // 将前k个数直接插入进multiset中，注意是小于K
            if(leastNumbers.size() < k)
            {
                leastNumbers.insert(*iter);
            }
            else
            {
                // 因为设置的从大到小排序，故multiset中第一个位置的元素即为最大值
                iterGreater = leastNumbers.begin();
                 
                // 如果input中当前元素比multiset中最大元素小，则替换；即保持multiset中这k个元素是最小的。
                if(*iter < *(leastNumbers.begin()))
                {
                    // 替换掉当前最大值
                    leastNumbers.erase(iterGreater);
                    leastNumbers.insert(*iter);
                }
            }
        }
         
        for(iterGreater = leastNumbers.begin();iterGreater!=leastNumbers.end();++iterGreater)
        {
            result.push_back(*iterGreater); // 将multiset中这k个元素输出
        }
         
        return result;
    }
};