Permutations II

最新推荐文章于 2020-03-29 17:21:57 发布

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本文探讨了在可能包含重复元素的整数集合上生成所有唯一排列的方法，并提供了两种优化解决方案，一种通过确保每次选择元素后的子数组有序来提高效率，另一种通过计数并依次取出每个数来避免重复，确保通过judgelarge测试。

Given a collection of numbers that might contain duplicates, return all possible unique permutations.

For example,
[1,1,2] have the following unique permutations:
[1,1,2], [1,2,1], and [2,1,1].

Failed on judge large.

public class Solution {
    public ArrayList<ArrayList<Integer>> res;
    public ArrayList<Integer> list;
	public ArrayList<ArrayList<Integer>> permuteUnique(int[] num) {
		// Start typing your Java solution below
		// DO NOT write main() function
		res = new ArrayList<ArrayList<Integer>>();
		list = new ArrayList<Integer>();
		perm(num, 0);
		return res;
	}
	public void perm(int[] num, int cur){
		if(cur == num.length){
			for(int i = 0; i < num.length; i++)
				list.add(num[i]);
            if(!res.contains(list))
			    res.add(list);
			list = new ArrayList<Integer>();
		}
		for(int i = cur; i < num.length; i++){
			int temp = num[i];
			num[i] = num[cur];
			num[cur] = temp;
			perm(num, cur + 1);
			temp = num[i];
			num[i] = num[cur];
			num[cur] = temp;
		}
	}
}

如果只是增加限定条件 if (i + 1 < num.length && num[i] == num[i + 1]) continue;

需要保证拿出num[i]之后，num[i....length - 1]部分也应该是排好序的，考虑到程序后面还要还原数组num，过于繁琐，此算法不适合改进。可以参考Jilong大神http://blog.youkuaiyun.com/sunjilong/article/details/8259986用此法。

下面方法统计每个数的个数然后依次取出，保证了取出数以后剩下数组中数的顺序。通过了judge large。参考自Tingmei大神。

public class Solution {
    public ArrayList<ArrayList<Integer>> res;
	public ArrayList<Integer> list;
	public int[] temp;
	public ArrayList<Integer> value;
	public ArrayList<Integer> counter;

	public ArrayList<ArrayList<Integer>> permuteUnique(int[] num) {
		// Start typing your Java solution below
		// DO NOT write main() function
		res = new ArrayList<ArrayList<Integer>>();
		list = new ArrayList<Integer>();
		value = new ArrayList<Integer>();
		counter = new ArrayList<Integer>();
		Arrays.sort(num);
		int val = num[0];
		int count = 1;
		for (int i = 1; i < num.length; i++) {
			if (val == num[i])
				count++;
			else{
				value.add(val);
				counter.add(count);
				val = num[i];
				count = 1;
			}
		}
		value.add(val);
		counter.add(count);
		temp = new int[num.length];
		perm(temp, 0);
		return res;
	}

	public void perm(int[] temp, int cur) {
		if (cur == temp.length) {
			for (int i = 0; i < temp.length; i++)
				list.add(temp[i]);
			res.add(list);
			list = new ArrayList<Integer>();
		} else {
			for (int i = 0; i < value.size(); i++) {
				if (counter.get(i) > 0){
					counter.set(i, counter.get(i) - 1);
					temp[cur] = value.get(i);
					perm(temp, cur + 1);
					counter.set(i, counter.get(i) + 1);
				}
			}
		}
	}
}