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NextPermutation.java
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71 lines (69 loc) · 2.52 KB
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/*
Author: King, wangjingui@outlook.com
Date: Dec 19, 2014
Problem: Next Permutation
Difficulty: Medium
Source: https://oj.leetcode.com/problems/next-permutation/
Notes:
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
Solution: O(n)
Processes: Take A = {1,3,2} as an example:
1. Traverse from back to forth, find the turning point, that is A[i] = 3.
2. Sort from the turning point to the end (A[i] to A[end]), so {3,2} becomes {2,3}.
3. If i equals to 0, finish! Else, goto 4.
4. Let j = i, search from A[j] to A[end] to find the first elem which is larger than A[i-1], '2' here.
5. Swap the elem A[j] with A[i-1].
Finally, the next permutation is {2,1,3}.
*/
public class Solution {
public void nextPermutation_1(int[] num) {
int last = num.length - 1;
int i = last;
while (i >0 && num[i - 1] >= num [i]) --i;
if (i == 0) {
for (int l = 0, r = last; l < r; ++l, --r) {
int tmp = num[l];
num[l] = num[r];
num[r] = tmp;
}
return;
}
for (int j = last; j >= i; --j) {
if (num[j] > num[i-1]) {
int tmp = num[j];
num[j] = num[i-1];
num[i-1] = tmp;
for (int l = i, r = last; l < r; ++l, --r) {
int t = num[l];
num[l] = num[r];
num[r] = t;
}
return;
}
}
}
public void nextPermutation_2(int[] num) {
int last = num.length - 1;
int i = last;
while (i > 0 && num[i - 1] >= num [i]) --i;
for (int l = i, r = last; l < r; ++l, --r) {
num[l] = num[l] ^ num[r];
num[r] = num[l] ^ num[r];
num[l] = num[l] ^ num[r];
}
if (i == 0) {
return;
}
int j = i;
while (j <= last && num[i-1] >= num[j]) ++j;
num[i-1] = num[i-1] ^ num[j];
num[j] = num[i-1] ^ num[j];
num[i-1] = num[i-1] ^ num[j];
}
}