LeetCode solution Report-Next Permutation

LeetCode solution Report-Next Permutation

Question:

Implement next permutation, which rearranges numbers into the lexicographically next greater permutation of numbers.
If such arrangement is not possible, it must rearrange it as the lowest possible order (ie, sorted in ascending order ).
The replacement must be in-place, do not allocate extra memory.
Here are some examples. Inputs are in the left-hand column and its corresponding outputs are in the right-hand column.
1, 2, 3 → 1, 3, 2
3, 2, 1 → 1, 2, 3
1, 1, 5 → 1, 5, 1
Click solution: Next Permutation

Analysis: finds the next arrangement of a given sequence. If not, finds the initial arrangement of the sequence.
Ideas:
The next sequence must be the smallest value-added value of the current sequence, that is, the question is converted to finding the smallest increment of the current sequence. It is the next series, as shown in:
Release/yaGjyc/release + release/release + qyry13bz1tfy0 + iwgwvrX49Kqx/release + DQo8cHJlIGNsYXNzPQ = "brush: java;">public class Solution { public void nextPermutation(int[] nums) { //Save the first index of nums[i] < nums[k] from right to left int i = 0; //Save the first index of nums[j] > nums[i] from right to left int j = 0; //To find the first index of nums[i] < nums[k] from right to left for(int k = nums.length - 2;k > 0;k --){ if(nums[k] < nums[k + 1]){ i = k; break; } } //To find the first index of nums[j] > nums[i] from right to left for(int l = nums.length - 1;l > 0;l --){ if(nums[l] > nums[i]){ j = l; break; } } //Swap nums[i] and nums[j] int swap = nums[i]; nums[i] = nums[j]; nums[j] = swap; //Reverse the sub array from i + 1 (or 0) to nums.length - 1 int n = 0; if(i == 0 && j == 0){ n = 0; }else{ n = i + 1; } int m = nums.length - 1; while(n < m){ int temp = nums[n]; nums[n] = nums[m]; nums[m] = temp; m --; n ++; } }}

Python code: Accepted

class Solution(object):    def nextPermutation(self, nums):                :type nums: List[int]        :rtype: void Do not return anything, modify nums in-place instead.                         i for the first index nums[k] > nums[k + 1] from right to left        j for the first index nums[i] < nums[j] from right to left                i = j = 0        #To find the the first index nums[k] > nums[k + 1] from right to left        k = len(nums) - 2        while(k > 0):            if(nums[k] < nums[k + 1]):                i = k                break            k -= 1        #To find the the first index nums[i] < nums[j] from right to left        k = len(nums) - 1        while(k > 0):            if(nums[k] > nums[i]):                j = k                break;            k -= 1        #Swap nums[i] and nums[j]            nums[i],nums[j] = nums[j],nums[i]        #Reverse elements for i + 1(or 0) to len(nums) - 1        if(i == 0 and j == 0):            n = 0        else:            n = i + 1        m = len(nums) - 1        while(n <= m):            nums[n],nums[m] = nums[m],nums[n]            n += 1            m -= 1