[leetcode]Permutations

Permutations

 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].

【吐来个槽】还以为要输出题目中给出的顺序，折腾了很久发现，你妹，坑爹呢！哼╭(╯^╰)╮

算法思路：

思路2：

dfs

代码如下：

public class Solution {
    List<List<Integer>> res = new ArrayList<List<Integer>>();
    public List<List<Integer>> permute(int[] num) {
        if(num == null || num.length == 0) return res;
        dfs(new ArrayList<Integer>(),num,new HashSet<Integer>());
        return res;
    }
    private void dfs(List<Integer> list,int[] num,Set<Integer> visited){
        if(list.size() == num.length){
            res.add(new ArrayList<Integer>(list));
            return;
        }
        for(int i = 0; i < num.length; i++){
            if(visited.contains(num[i])) continue;
            visited.add(num[i]);
            list.add(num[i]);
            dfs(list,num,visited);
            visited.remove(num[i]);
            list.remove(list.size() - 1);
        }
    }
}

思路1：

迭代，先求出(n - 1)的排列情况，继而求出n的。

public class Solution {
    public List<List<Integer>> permute(int[] num) {
        List<List<Integer>> result = new ArrayList<List<Integer>>();
        if(num == null || num.length == 0) return result;
        if(num.length == 1){
            List<Integer> list = new ArrayList<Integer>();
            list.add(num[0]);
            result.add(list);
            return result;
        }
        int[] pre = Arrays.copyOf(num, num.length - 1);
        int tem = num[num.length - 1];
        for(List<Integer> list : permute(pre)){
            int newLength = list.size() + 1;
            for(int i = 0; i < newLength; i++){
                List<Integer> newList = new ArrayList<Integer>();
                for(int j = 0; j < i;newList.add(list.get(j++)));
                newList.add(tem);
                for(int j = i; j < newLength - 1;newList.add(list.get(j++)));
                result.add(new ArrayList<Integer>(newList));
            }
        }
        return result;
    }
}