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LeetCode-90-Subsets II(回溯)-Medium

題意理解:

同題78,但所給數列中可能存在重複元素;

題目分析:

1. 解題思路同題78;

2. 在同級遍歷中,如果發現當前元素在之前已出現,則跳過;

解題程式碼:

public class Solution {
    private ArrayList<ArrayList<Integer>> ans= new ArrayList<ArrayList<Integer>>();
    
    private boolean find(int[] nums, int indexBegin, int indexCur, int n){
        if(indexCur==indexBegin){
            return false;
        }
        
        boolean ans=false;
        for(int i=indexBegin; i<indexCur; i++){
            if(nums[i]==n){
                ans=true;
                break;
            }
        }
        
        return ans;
    }
    
    private void process(ArrayList<Integer> list, int[] nums, int initPos, int n){
        if(n==0){
            ans.add(list);
            return;
        }
        
        if(initPos==nums.length){
            return;
        }
        
        for(int i=initPos; i<nums.length; i++){
            if(find(nums, initPos, i, nums[i])){
                continue;
            }
            ArrayList<Integer> tList= new ArrayList(list);
            tList.add(new Integer(nums[i]));
            process(tList, nums, i+1, n-1);
        }
    }
    
    public List<List<Integer>> subsetsWithDup(int[] nums) {
        Arrays.sort(nums);
        
        for(int i=0; i<=nums.length; i++){
            ArrayList<Integer> list=new ArrayList<Integer>();
            process(list, nums, 0, i);
        }
        
        return (List)ans;
    }
}