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Permutations.java
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/*
Author: King, wangjingui@outlook.com
Date: Dec 25, 2014
Problem: Permutations
Difficulty: Easy
Source: https://oj.leetcode.com/problems/permutations/
Notes:
Given a collection of numbers, return all possible permutations.
For example,
[1,2,3] have the following permutations:
[1,2,3], [1,3,2], [2,1,3], [2,3,1], [3,1,2], and [3,2,1].
Solution: dfs...
*/
public class Solution {
public List<List<Integer>> permute_1(int[] num) {
List<List<Integer>> res = new ArrayList<List<Integer>>();
List<Integer> path = new ArrayList<Integer>();
boolean[] free = new boolean[num.length];
Arrays.fill(free, true);
permuteRe(num, res, path,free);
return res;
}
void permuteRe(int[] num, List<List<Integer>> res, List<Integer> path, boolean[] free) {
if(path.size() == num.length) {
ArrayList<Integer> tmp = new ArrayList<Integer>(path);
res.add(tmp);
return;
}
for (int i = 0; i < num.length; ++i) {
if (free[i] == true) {
free[i] = false;
path.add(num[i]);
permuteRe(num, res, path, free);
path.remove(path.size() - 1);
free[i] = true;
}
}
}
public boolean nextPermutation(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 false;
}
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];
return true;
}
public List<List<Integer>> permute_2(int[] num) {
List<List<Integer>> res = new ArrayList<List<Integer>>();
Arrays.sort(num);
do {
List<Integer> path = new ArrayList<Integer>();
for (int i : num) path.add(i);
res.add(path);
} while(nextPermutation(num));
return res;
}
}