g0501_0600.s0556_next_greater_element_iii.Solution Maven / Gradle / Ivy
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Java Solution for LeetCode algorithm problems, continually updating
package g0501_0600.s0556_next_greater_element_iii;
// #Medium #String #Math #Two_Pointers #Programming_Skills_II_Day_10
// #2022_08_03_Time_0_ms_(100.00%)_Space_39.2_MB_(89.26%)
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--;
}
}
}