标签：
Given a collection of distinct 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].
解法一：
the basic idea is, to permute n numbers, we can add the nth number into the resultingList<List<Integer>> from the n-1 numbers, in every possible position.
For example, if the input num[] is {1,2,3}: First, add 1 into the initial List<List<Integer>> (let‘s call it "answer").
Then, 2 can be added in front or after 1. So we have to copy the List in answer (it‘s just {1}), add 2 in position 0 of {1}, then copy the original {1} again, and add 2 in position 1. Now we have an answer of {{2,1},{1,2}}. There are 2 lists in the current answer.
Then we have to add 3. first copy {2,1} and {1,2}, add 3 in position 0; then copy {2,1} and {1,2}, and add 3 into position 1, then do the same thing for position 3. Finally we have 2*3=6 lists in answer, which is what we want.
public List<List<Integer>> permute(int[] num) { List<List<Integer>> ans = new ArrayList<List<Integer>>(); if (num.length ==0) return ans; List<Integer> l0 = new ArrayList<Integer>(); l0.add(num[0]); ans.add(l0); for (int i = 1; i< num.length; ++i){ List<List<Integer>> new_ans = new ArrayList<List<Integer>>(); for (int j = 0; j<=i; ++j){ for (List<Integer> l : ans){ List<Integer> new_l = new ArrayList<Integer>(l); new_l.add(j,num[i]); new_ans.add(new_l); } } ans = new_ans; } return ans; }
reference：https://leetcode.com/discuss/19510/my-ac-simple-iterative-java-python-solution
解法二： 九章的解法，不太明白
public class Solution { public ArrayList<ArrayList<Integer>> permute(int[] num) { ArrayList<ArrayList<Integer>> rst = new ArrayList<ArrayList<Integer>>(); if (num == null || num.length == 0) { return rst; } ArrayList<Integer> list = new ArrayList<Integer>(); helper(rst, list, num); return rst; } public void helper(ArrayList<ArrayList<Integer>> rst, ArrayList<Integer> list, int[] num){ if(list.size() == num.length) { rst.add(new ArrayList<Integer>(list)); return; } for(int i = 0; i<num.length; i++){ if(list.contains(num[i])){ continue; } list.add(num[i]); helper(rst, list, num); list.remove(list.size() - 1); } } }
// Non-Recursion class Solution { /** * @param nums: A list of integers. * @return: A list of permutations. */ public ArrayList<ArrayList<Integer>> permute(ArrayList<Integer> nums) { ArrayList<ArrayList<Integer>> permutations = new ArrayList<ArrayList<Integer>>(); if (nums == null || nums.size() == 0) { return permutations; } int n = nums.size(); ArrayList<Integer> stack = new ArrayList<Integer>(); stack.add(-1); while (stack.size() != 0) { Integer last = stack.get(stack.size() - 1); stack.remove(stack.size() - 1); // increase the last number int next = -1; for (int i = last + 1; i < n; i++) { if (!stack.contains(i)) { next = i; break; } } if (next == -1) { continue; } // generate the next permutation stack.add(next); for (int i = 0; i < n; i++) { if (!stack.contains(i)) { stack.add(i); } } // copy to permutations set ArrayList<Integer> permutation = new ArrayList<Integer>(); for (int i = 0; i < n; i++) { permutation.add(nums.get(stack.get(i))); } permutations.add(permutation); } return permutations; } }
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原文地址：http://www.cnblogs.com/hygeia/p/5094347.html
