輸入n個整數，找出其中最小的K個數。例如輸入4,5,1,6,2,7,3,8這8個數字，則最小的4個數字是1,2,3,4。

解題思路

• 排序後遍歷（相當於簡化後的暴力）O(logn)
• 藉助快排的Partition思想O(n)

上程式碼（C++香）

法一：排序後遍歷（相當於簡化後的暴力）

class Solution {
public:
    void mySwap(vector<int> &num, int i, int j){
        int temp = num[j];
        num[j] = num[i];
        num[i] = temp;
    }

    int myPartition(vector<int> &num, int low, int high){
        int pivot = num[low];
        while(low < high){
            // 將右邊比pivot小的放到左邊
            while(low < high && num[high] >= pivot)
                high--;
            mySwap(num, low, high);
            while(low < high && num[low] <= pivot)
                low++;
            mySwap(num, low, high);
        }
        return low;
    }

    // 快排
    void QSort(vector<int> &num, int low, int high){
        if(low >= high)
            return ;
        int pivot = myPartition(num, low, high);
        QSort(num, low, pivot - 1);
        QSort(num, pivot + 1, high);
    }

    vector<int> GetLeastNumbers_Solution(vector<int> input, int k) {
        vector<int> ans;
        if(k > input.size())
            return ans;
        QSort(input, 0, input.size() - 1);
        for(int i = 0; i < k; i++)
            ans.push_back(input[i]);
        return ans;
    }
};
 

法二：藉助快排的Partition思想

  得到pivot如果與k-1相等，那麼陣列左邊的數已經就比num[k-1]小了，所以找到這個pivot就行。

void mySwap(vector<int> &num, int i, int j){
    int temp = num[j];
    num[j] = num[i];
    num[i] = temp;
}

int myPartition(vector<int> &num, int low, int high){
    int pivot = num[low];
    while(low < high){
        // 將右邊比pivot小的放到左邊
        while(low < high && num[high] >= pivot)
            high--;
        mySwap(num, low, high);
        while(low < high && num[low] <= pivot)
            low++;
        mySwap(num, low, high);
    }
    return low;
}

vector<int> GetLeastNumbers_Solution(vector<int> input, int k) {
    vector<int> ans;
    int length = input.size();
    if(k > length)
        return ans;
    int low = 0;
    int high = length - 1;
    int pivot = myPartition(input, low, high);
    while(pivot != k - 1){
        // k-1在左邊繼續找
        if(pivot > k - 1){
            high = pivot - 1;
            pivot = myPartition(input, low, high);
        }
        else{
            low = pivot + 1;
            pivot = myPartition(input, low, high);
        }
    }

    for(int i = 0; i < k; i++)
        ans.push_back(input[i]);
    return ans;
}