寻找第i 小的数

快速选择算法解析

思想借鉴快排, 但是时间复杂度是快排的一半

public class test1 {

    public static int serarch(int[] array, int low, int high, int findIndex){
        int lowHis = low;
        int highHis = high;

        if(low >= high) return array[low];
        int temp = array[low];

        while (low < high){
            while (high > low  && array[high] >= temp){
                high = high - 1;
            }
            array[low] = array[high];
            while (low < high && array[low] <= temp){
                low = low + 1;
            }
            array[high] = array[low];
        }
        array[low] = temp;

        if(findIndex <= low){
            return serarch(array, lowHis, low, findIndex);
        } else{
            return serarch(array, low + 1, highHis, findIndex);
        }
    }

    public static void main(String args[]){
        int[] array = {6, 4, 5, 3, 9, 8, 3, 10, 1, 7};
        System.out.println(serarch(array, 0, array.length - 1, 2));
    }
}

 

以下为几种用C语言实现寻找第K小的的方法。 #### 基于快速排序思想的实现 ```c #include <stdio.h> // 交换两个元素 void swap(int *a, int *b) { int temp = *a; *a = *b; *b = temp; } // 分区函 int partition(int arr[], int low, int high) { int pivot = arr[high]; int i = (low - 1); for (int j = low; j <= high - 1; j++) { if (arr[j] < pivot) { i++; swap(&arr[i], &arr[j]); } } swap(&arr[i + 1], &arr[high]); return (i + 1); } // 寻找第K小的 int findKthSmallest(int arr[], int low, int high, int k) { if (k > 0 && k <= high - low + 1) { int index = partition(arr, low, high); if (index - low == k - 1) return arr[index]; if (index - low > k - 1) return findKthSmallest(arr, low, index - 1, k); return findKthSmallest(arr, index + 1, high, k - index + low - 1); } return -1; } int main() { int arr[] = {7, 10, 4, 3, 20, 15}; int n = sizeof(arr) / sizeof(arr[0]); int k = 3; int result = findKthSmallest(arr, 0, n - 1, k); if (result != -1) printf("第 %d 小的是: %d\n", k, result); else printf("无效的K值\n"); return 0; } ``` #### 另一种基于快速排序思想的实现(参考已有引用) ```c //寻找第k小元素 int select_k(int *a, int low, int high, int k) { int p, q, i, mm; //mm为中项集合的中项 int *M = (int *)malloc(N * sizeof(int)); int *temp = (int *)malloc(N * sizeof(int)); //辅助组 p = high - low + 1; //元素总个 if (p < 6) //若p小于阈值44直接排序得到第k小元素 { Mergesort(a, low, high, temp); return a[k]; } q = p / 5; //将所有元素分成的总组 for (i = 1; i <= q; i++) { Mergesort(a, low + 5 * (i - 1), low + 5 * (i - 1) + 4, temp); M[i] = a[low + 5 * (i - 1) + 2]; } if (q == 1) { mm = M[1]; } else mm = select_k(M, 1, q, q / 2); int count1 = 1, count2 = 1, count3 = 1; int *A1 = (int *)malloc(N * sizeof(int)); int *A2 = (int *)malloc(N * sizeof(int)); int *A3 = (int *)malloc(N * sizeof(int)); for (i = low; i <= high; i++) { if (a[i] < mm) { A1[count1++] = a[i]; } else if (a[i] == mm) { A2[count2++] = a[i]; } else { A3[count3++] = a[i]; } } if (count1 - 1 >= k) { return select_k(A1, 1, count1 - 1, k); } else if (count1 - 1 + count2 - 1 >= k) { return mm; } else if (count1 - 1 + count2 - 1 < k) { return select_k(A3, 1, count3 - 1, k - (count1 - 1) - (count2 - 1)); } else return 0; } ``` 注:上述代码中的`Mergesort`函未给出实现,需要自行实现归并排序函
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