leetcode Permutations

最新推荐文章于 2019-05-10 11:15:43 发布

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本文详细解析了LeetCode上的排列问题，提供了AC代码示例。通过递归算法生成所有可能的排列组合，并针对Java中数组复制的问题进行了特殊处理。

Permutations 原题地址：

https://oj.leetcode.com/problems/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].

这道题解法很多，http://blog.youkuaiyun.com/tuantuanls/article/details/8717262

这里由于涉及到java数组复制只是shadow clone的原因，代码略烦，贴一段比较简单而且有助于解决Permutations II的代码，已AC。

public class Solution {
    List<List<Integer>> list = new ArrayList<List<Integer>>();
	public List<List<Integer>> permute(int[] num) {
		if (num == null || num.length == 0)
			return list;
		boolean[] check = new boolean[num.length + 1];
		LinkedList<Integer> temp = new LinkedList<Integer>();
		permutelist(num, check, temp);
		return list;
	}
	
	private void permutelist(int[] num, boolean[] check, LinkedList<Integer> temp) {
		if (temp.size() == num.length) {
			LinkedList<Integer> plist = (LinkedList<Integer>) temp.clone();
			list.add(plist);
		}
		for (int i = 0; i < num.length; i++) {
			if (!check[i]) {
				temp.add(num[i]);
				check[i] = true;
				permutelist(num, check, temp);
				check[i] = false;
				temp.pollLast();
			}
		}
	}
}public class Solution {
    List<List<Integer>> list = new ArrayList<List<Integer>>();
	public List<List<Integer>> permute(int[] num) {
		if (num == null || num.length == 0)
			return list;
		boolean[] check = new boolean[num.length + 1];
		LinkedList<Integer> temp = new LinkedList<Integer>();
		permutelist(num, check, temp);
		return list;
	}
	
	private void permutelist(int[] num, boolean[] check, LinkedList<Integer> temp) {
		if (temp.size() == num.length) {
			LinkedList<Integer> plist = (LinkedList<Integer>) temp.clone();
			list.add(plist);
		}
		for (int i = 0; i < num.length; i++) {
			if (!check[i]) {
				temp.add(num[i]);
				check[i] = true;
				permutelist(num, check, temp);
				check[i] = false;
				temp.pollLast();
			}
		}
	}
}