# The Problem:

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

# Notes:

- A tricky question that I think is too hard for an interview if haven’t met it before, it’s a classic one though.
- Steps:
- For an always descending array, it’s simple that only need to sort the array in ascending order.
- Otherwise means It has ascending pair somewhere, find the last index that with a larger number following. Note that beyond that “last index”, the array is in descending order. Then reverse the descending half to make it in ascending order, and swap the “last index” number with the smallest one that is greater than it in the latter half.

# Java Solution

public class Solution { public void nextPermutation(int[] num) { if(num.length <=1)return; int indexInc = -1; for(int i = 1; i< num.length; i++){ if(num[i] > num[i-1]){ indexInc = i-1; } } if(indexInc ==-1){ // decreaseing all the way Arrays.sort(num); return; } else{ int i = indexInc+1; int j = num.length-1; while(i<j){ int temp = num[i]; num[i] = num[j]; num[j] = temp; i++; j--; } for(int k = indexInc+1; k<num.length; k++){ if(num[k] > num[indexInc]){ int temp = num[k]; num[k] = num[indexInc]; num[indexInc] = temp; break; } } return; } } }

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