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].
思路：生成permutation，这题假设没有重复元素，递归求解，每一次在cur位置填元素，添之前判断钙元素是否已经用过，没用过继续下去，指导cur==n。
public class Solution { public ArrayList > permute(int[] num) { if (num == null) return null; ArrayList > res = new ArrayList >(); if (num.length == 0) return res; int[] permSeq = new int[num.length]; perm(num.length, 0, num, permSeq, res); return res; } private void perm(int n, int cur, int[] num, int[] perm, ArrayList > res) { if (cur == n) { ArrayList tmp = new ArrayList (); for (int i = 0; i < perm.length; i++) { tmp.add(perm[i]); } res.add(tmp); } else { int i; for (i = 0; i < num.length; i++) { int j; boolean ok = true; for (j = cur - 1; j >= 0; j--) { if (perm[j] == num[i]) ok = false; } if (ok) { perm[cur] = num[i]; perm(n, cur + 1, num, perm, res); } } } } public static void main(String[] args) { System.out.println(new Solution().permute(new int[] { 1, 2, 3,4 })); }}