LeetCode-in-Java
556. Next Greater Element III
Medium
Given a positive integer n, find the smallest integer which has exactly the same digits existing in the integer n and is greater in value than n. If no such positive integer exists, return -1.
Note that the returned integer should fit in 32-bit integer, if there is a valid answer but it does not fit in 32-bit integer, return -1.
Example 1:
Input: n = 12
Output: 21
Example 2:
Input: n = 21
Output: -1
Constraints:
1 <= n <= 231 - 1
Solution
public class Solution {
/*
- What this problem wants is finding the next permutation of n
- Steps to find the next permuation:
find largest index k such that inp[k] < inp[k+1];
if k == -1: return -1
else:
look for largest index l such that inp[l] > inp[k]
swap the two index
reverse from k+1 to n.length
*/
public int nextGreaterElement(int n) {
char[] inp = String.valueOf(n).toCharArray();
// Find k
int k = -1;
for (int i = inp.length - 2; i >= 0; i--) {
if (inp[i] < inp[i + 1]) {
k = i;
break;
}
}
if (k == -1) {
return -1;
}
// Find l
int largerIdx = inp.length - 1;
for (int i = inp.length - 1; i >= 0; i--) {
if (inp[i] > inp[k]) {
largerIdx = i;
break;
}
}
swap(inp, k, largerIdx);
reverse(inp, k + 1, inp.length - 1);
// Build result
int ret = 0;
for (char c : inp) {
int digit = c - '0';
// Handle the case if ret > Integer.MAX_VALUE - This idea is borrowed from problem 8.
// String to Integer (atoi)
if (ret > Integer.MAX_VALUE / 10
|| (ret == Integer.MAX_VALUE / 10 && digit > Integer.MAX_VALUE % 10)) {
return -1;
}
ret = ret * 10 + (c - '0');
}
return ret;
}
private void swap(char[] inp, int i, int j) {
char temp = inp[i];
inp[i] = inp[j];
inp[j] = temp;
}
private void reverse(char[] inp, int start, int end) {
while (start < end) {
char temp = inp[start];
inp[start] = inp[end];
inp[end] = temp;
start++;
end--;
}
}
}