# Java Solution with in-line explanation

• The same problem with https://leetcode.com/problems/next-permutation

``````public class Solution {
public int nextGreaterElement(int n) {
char[] chars = (n + "").toCharArray();

int l = chars.length;
if (l < 2) return -1;
int[] nums = new int[l];

for (int i = 0; i < l; i++) nums[i] = chars[i] - '0';

//Start from its last element, traverse backward to find the first one with index i that satisfy
// nums[i-1] < nums[i]. So, elements from nums[i] to nums[l-1] is reversely sorted.
int index = l - 1;
while (index > 0) {
if (nums[index - 1] < nums[index]) break;
index--;
}

//To find the next permutation, we have to swap some numbers at different positions,
//to minimize the increased amount, we have to make the highest changed position
// as high as possible. Notice that index larger than or equal to i is not possible as
// nums[i,l-1] is reversely sorted. So, we want to increase the number at index i-1,
// clearly, swap it with the smallest number between nums[i,l-1] that is larger than nums[i-1].
// For example, original number is 121543321, we want to swap the '1' at position 2 with '2' at position 7.
if (index == 0) {
return -1;
}
else {
//The last step is to make the remaining higher position part as small as possible,
// we just have to reversely sort the nums[i,l-1]
int val = nums[index - 1];
int j = l - 1;
while (j >= index){
if (nums[j] > val) break;
j--;
}
swap(nums, j, index - 1);

reverse(nums, index, l - 1);
}

long result = 0;
for (int i = 0; i < l; i++) {
result = result * 10 + nums[i];
}

return result <= Integer.MAX_VALUE ? (int)result : -1;
}

public void swap(int[] nums, int i, int j){
int temp = nums[i];
nums[i] = nums[j];
nums[j] = temp;
}

public void reverse(int[] nums, int start, int end){
if (start > end) return;
for (int i = start; i <= (end + start) / 2; i++)
swap(nums, i, start + end - i);
}
}
``````

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