[leetcode] Permutations II

标签：class   code   java   http   tar   com   

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].

https://oj.leetcode.com/problems/permutations-ii/

思 路：有重复元素的数组生成permutation，要去除掉重复的情况：先排序，然后还是permutation的思路，依次往cur位置填元素，因为有 些元素有多次，所以需要用c1,c2来统计已经填好的数量和总共需要填的数量，并且因为排序了，相同元素的去重只需要跟前一个元素比较一下是否相等 （if (i == 0 || num[i] != num[i - 1])）。

import java.util.ArrayList;
import java.util.Arrays;

public class Solution {
	public ArrayList<ArrayList<Integer>> permuteUnique(int[] num) {
		if (num == null)
			return null;
		ArrayList<ArrayList<Integer>> res = new ArrayList<ArrayList<Integer>>();
		if (num.length == 0)
			return res;
		Arrays.sort(num);
		int[] permSeq = new int[num.length];
		perm(num.length, 0, num, permSeq, res);
		return res;
	}

	private void perm(int n, int cur, int[] num, int[] perm,
			ArrayList<ArrayList<Integer>> res) {
		if (cur == n) {
			ArrayList<Integer> tmp = new ArrayList<Integer>();
			for (int i = 0; i < perm.length; i++) {
				tmp.add(perm[i]);
			}
			res.add(tmp);
		} else {
			int i;
			for (i = 0; i < num.length; i++)
				// this "if" is the key part
				if (i == 0 || num[i] != num[i - 1]) {
					int j;
					int c1 = 0, c2 = 0;
					for (j = 0; j < num.length; j++) {
						if (num[i] == num[j])
							c1++;
					}

					for (j = cur - 1; j >= 0; j--) {
						if (perm[j] == num[i])
							c2++;
					}
					if (c2 < c1) {
						perm[cur] = num[i];
						perm(n, cur + 1, num, perm, res);
					}
				}
		}

	}

	public static void main(String[] args) {
		System.out.println(new Solution().permuteUnique(new int[] { -1, 2, -1,
				2, 1, -1, 2, 1 }));
	}
}

[leetcode] Permutations II,布布扣,bubuko.com

[leetcode] Permutations II

标签：class   code   java   http   tar   com   

原文地址：http://www.cnblogs.com/jdflyfly/p/3810756.html