力扣-46. 全排列

0.题目

1.飞哥的题解

class Solution {
    public List<List<Integer>> permute(int[] nums) {
        List<Integer> numList = new ArrayList<>();
        for (int i = 0; i < nums.length; i++) {
            numList.add(nums[i]);
        }
        List<List<Integer>> resultList = new ArrayList<>();
        List<Integer> rightList = new ArrayList<>();
        find(resultList, rightList, numList);
        return resultList;
    }

    private void find(List<List<Integer>> resultList, List<Integer> rightList, List<Integer> numList) {
        for (int i = 0; i < numList.size(); i++) {
            Integer number = numList.remove(i);
            rightList.add(number);
            if (numList.isEmpty()) {
                resultList.add(new ArrayList<>(rightList));
                numList.add(number);
                rightList.remove(rightList.size()-1);
                return;
            }
            find(resultList, rightList, numList);
            numList.add(i,number);
            rightList.remove(rightList.size()-1);
        }
    }
}

2.官方题解

class Solution {
    public List<List<Integer>> permute(int[] nums) {
        List<List<Integer>> res = new ArrayList<List<Integer>>();

        List<Integer> output = new ArrayList<Integer>();
        for (int num : nums) {
            output.add(num);
        }

        int n = nums.length;
        backtrack(n, output, res, 0);
        return res;
    }

    public void backtrack(int n, List<Integer> output, List<List<Integer>> res, int first) {
        // 所有数都填完了
        if (first == n) {
            res.add(new ArrayList<Integer>(output));
        }
        for (int i = first; i < n; i++) {
            // 动态维护数组
            Collections.swap(output, first, i);
            // 继续递归填下一个数
            backtrack(n, output, res, first + 1);
            // 撤销操作
            Collections.swap(output, first, i);
        }
    }
}

作者：LeetCode-Solution
链接：https://leetcode-cn.com/problems/permutations/solution/quan-pai-lie-by-leetcode-solution-2/
来源：力扣（LeetCode）
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。

class Solution {
public:
    void backtrack(vector<vector<int>>& res, vector<int>& output, int first, int len){
        // 所有数都填完了
        if (first == len) {
            res.emplace_back(output);
            return;
        }
        for (int i = first; i < len; ++i) {
            // 动态维护数组
            swap(output[i], output[first]);
            // 继续递归填下一个数
            backtrack(res, output, first + 1, len);
            // 撤销操作
            swap(output[i], output[first]);
        }
    }
    vector<vector<int>> permute(vector<int>& nums) {
        vector<vector<int> > res;
        backtrack(res, nums, 0, (int)nums.size());
        return res;
    }
};

作者：LeetCode-Solution
链接：https://leetcode-cn.com/problems/permutations/solution/quan-pai-lie-by-leetcode-solution-2/
来源：力扣（LeetCode）
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。

来源：oschina

链接：https://my.oschina.net/hengbao666/blog/4604791