Hey Coders! π Today, let's dive into the Java code for the 'Next Permutation' algorithm β a handy tool for generating the next lexicographically greater permutation of an array.
public class Solution {
public int[] nextPermutation(int[] A) {
// Code Magic Unleashed
}
}
Checking the Current Sequence:
We kick off by scanning the array to see if it's already in ascending order, indicating room for a greater permutation.
int count = 0;
for (int i = 1; i < A.length; i++) {
if (A[i] > A[i - 1]) {
count++;
break;
}
}
Rearranging for a Fresh Start:
If the array is already at its peak permutation, we perform a straightforward reversal to reset the order.
if (count == 0) {
int i = 0;
int j = A.length - 1;
while (i < j) {
int temp = A[i];
A[i] = A[j];
A[j] = temp;
i++;
j--;
}
return A; // Done and Dusted!
}
Identifying the Pivot:
Now, we identify a pivot point where a change in the permutation is needed.
int a = 0;
for (int i = A.length - 1; i > 0; i--) {
if (A[i - 1] < A[i]) {
a = i;
break;
}
}
Swapping to Form a New Sequence:
In the heart of the algorithm, we execute a swap at the pivot point, ensuring the formation of a new lexicographically greater permutation.
for (int i = A.length - 1; i > a - 1; i--) {
if (A[i] > A[a - 1]) {
int temp = A[a - 1];
A[a - 1] = A[i];
A[i] = temp;
break;
}
}
Final Exchange for Completion:
The last step involves a final exchange beyond the pivot point, completing the generation of the next permutation.
int b = A.length - 1;
while (a < b) {
int temp1 = A[a];
A[a] = A[b];
A[b] = temp1;
a++;
b--;
}
return A; // The End
In Summary:
This Java code elegantly navigates the complexities of generating the next lexicographically greater permutation. Understanding these steps opens the door to manipulating arrays efficiently in various applications.
#NextPermutation #JavaAlgorithm
Happy coding! ππ©βπ»
Jordan ^_~