[recursion, Java intelligence podcast notes

Cxy happy 2021-08-08 21:24:55
recursion java intelligence podcast notes


There will be n Array of Numbers arr Make a full arrangement , The idea used is recursion and backtracking .

  1. Yes n All elements are arranged , Swap the first element with the following elements in turn , Identify the first element

  2. Yes, later n-1 All elements are arranged ,**( It can be seen as a sub problem of the first step )** Recursive implementation

  3. Replace the exchanged elements back , To prevent the order of array elements from being disturbed **( Back to the mind )**

See the following functions for specific implementation ,( You can use it directly )


/**
* Arrange all the elements in the array
* @param arr Array to be arranged
* @param k Determine the number of elements , It's a subscript , from 0 Start
* */
private static void f(int[] arr, int k) {
// When k Equal to the length of the array , Explain that the arrangement is complete
if (k == arr.length) {
// Output the arranged array
for (int i = 0; i < arr.length; i++) {
System.out.print(arr[i]);
}
System.out.println();
}
for (int i = k; i < arr.length; i++) {
// Swap the position of the element to be determined with the subsequent element
int t = arr[k];
arr[k] = arr[i];
arr[i] = t;
// recursive ( Make sure No k+1 Elements )
f(arr, k+1);
// to flash back , Replace the replaced elements back
t = arr[k];
arr[k] = arr[i];
arr[i] = t;
}
}

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The test case :


public static void main(String[] args) {
int[] arr = {1,2,3,4}; // Array to be processed
int n = 3; // Take out the number of elements
int[] newarr = new int[n]; // An array of results
f(arr, 0);
}

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3、 ... and 、 Arrangement and combination of array elements

===============

With the above pair from n Array of elements arr Remove from m Number ( Do not consider the order and do not repeat ) And for n The number is fully arranged , So for the n Take out... From the number m The problem of realizing the arrangement of numbers , It can be seen as a combination of the above two problems .

According to the thinking in Mathematics , We can start from n Select from an array of elements m Elements , Then on this m All elements can be arranged in full order .

The implementation method is as follows :


/**
* Pair in array n The number is arranged completely
* @param Array to be processed
* @param newarr Array after arrangement
* @param k Which subscript element to start processing
* @param n Number of processing elements
* */
private static void pac(int[] arr,int[] newarr, int k,int n) {
// When n=0 when , Note the number of selected numbers is 0, That is, the combination completes
if (n==0) {
f(newarr, 0); // Fully arrange the new array combined
return;
}
for (int i = k; i <= arr.length-n; i++) {
newarr[newarr.length-n] = arr[i];
pac(arr, newarr,i+1, n-1);
}
}
/**
* Arrange all the elements in the array
* @param arr Array to be arranged
* @param k Determine the number of elements , It's a subscript , from 0 Start
* */
private static void f(int[] arr, int k) {
// When k Equal to the length of the array , Explain that the arrangement is complete
if (k == arr.length) {
// Output the arranged array
for (int i = 0; i < arr.length; i++) {
System.out.print(arr[i]);
}
System.out.println();
}
for (int i = k; i < arr.length; i++) {
// Swap the position of the element to be determined with the subsequent element
int t = arr[k];
arr[k] = arr[i];
arr[i] = t;
// recursive ( Make sure No k+1 Elements )
f(arr, k+1);
// to flash back , Replace the replaced elements back
t = arr[k];
arr[k] = arr[i];
arr[i] = t;
}
}

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The test case :

Last

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版权声明
本文为[Cxy happy]所创,转载请带上原文链接,感谢
https://javamana.com/2021/08/20210808212406165a.html

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