"Base algorithm" sort-complex sort of two (find out the number of k large)

The idea of cutting is the most essence of high-speed sequencing. Each cut out of the element K one ranked in the K-bit, so take advantage of this idea we know at least 3 points

1. The cut-out element K must finally be ranked in the K position.

2. The element to the left of K must be smaller or equal than K.

3. The element on the right of K must be larger or equal than K.

So we are able to navigate through these properties to an arbitrary element.

For example, after we partition an array, we get a={5,3,4,2,6,8,10,12,11,9}

A[k]=8, so we know that after the orderly a[5]=8, a[4] must be on the left of 8, a[6] must be on the 8 right

So, we must know that the number 8 is the 5+1 small number in the array. 第10-5 a large number

So we come to the assumption that the number a[k]=x is cut, then X must be the k+1 bit in the array, that is, the k+1 small number.

Assuming that the length of the array is n, then x is the number of n-k large in the array.

Here is the code for the cut

public static int partition (int[] array, int left, int. right) {int i = Left;int J = right + 1;while (true) {while (the more (AR Ray[left], Array[++i])) if (i = = right) Break;while (more (Array[--j], Array[left])) if (j = = left) break;if (i >= j) break;e Xchange (Array, I, j);} Exchange (Array, left, j); return J;}

How do you locate the other elements after cutting?

Suppose we locate the a[k]=x and find that the target element o is larger than x, then find it on the right, left=k+1, assuming that it is smaller than x, then find it on the left. Right=k-1, otherwise the location is successful.

public static Int. Select (int[] array, int k) {int left = 0;int right = Array.length-1;while (left < right) {int J = P Artition (array, left, right); if (J < k) left = j + 1;else if (J > k) right = J-1;elsereturn array[k];} return array[k];}

The following gives the complete code, only for everyone to participate in the test

Comparepublic static Boolean more (int v, int w) {return v > w;} Exchangepublic static void Exchange (int[] array, int i, int j) {int temp = Array[i];array[i] = array[j];array[j] = Temp ;} public static int partition (int[] array, int left, int. right) {int i = Left;int J = right + 1;while (true) {while (the more (AR Ray[left], Array[++i])) if (i = = right) Break;while (more (Array[--j], Array[left])) if (j = = left) break;if (i >= j) break;e Xchange (Array, I, j);} Exchange (Array, left, j); return J;} public static Int. Select (int[] array, int k) {int left = 0;int right = Array.length-1;while (left < right) {int J = P Artition (array, left, right); if (J < k) left = j + 1;else if (J > k) right = J-1;elsereturn array[k];} return array[k];}

"Base algorithm" sort-complex sort of two (find out the number of k large)