穷举vs暴搜vs深搜vs回溯vs剪枝专题一＞全排列II

穷举vs暴搜vs深搜vs回溯vs剪枝专题一＞全排列II

题目：

解析： 

这题设计递归函数，主要把看如何剪枝 

代码： 

class Solution {
    private List<List<Integer>> ret;
    private List<Integer> path;
    private boolean[] check;
    public List<List<Integer>> permuteUnique(int[] nums) {
        ret = new ArrayList<>();
        path = new ArrayList<>();
        check = new boolean[nums.length];
        Arrays.sort(nums);//

        dfs(nums,0);
        return ret;
    }

    //pos是决策树层数
    private void dfs(int[] nums,int pos){
        if(pos == nums.length){
            ret.add(new ArrayList<>(path));
            return;
        }

        for(int i = 0; i < nums.length; i++){ 
            
            //不合法分支剪枝
            if(check[i] == true || i != 0 && nums[i] == nums[i-1] && check[i-1] == false)
                continue;

            if(check[i] == false){
            path.add(nums[i]);
            check[i] = true;
            dfs(nums,pos+1);

            //恢复现场
            path.remove(path.size()-1);
            check[i] = false;
            }

        }
    }
}