47.Permutations-II
47. Permutations II
题目地址
https://leetcode.com/problems/permutations-ii/
https://www.lintcode.com/problem/permutations-ii/description
题目描述
Given a collection of numbers that might contain duplicates, return all possible unique permutations.
Example:
Input: [1,1,2]
Output:
[
[1,1,2],
[1,2,1],
[2,1,1]
]
代码
public class Solution {
public List<List<Integer>> permuteUnique(int[] nums) {
List<List<Integer>> results = new ArrayList<>();
if (nums == null) return results;
Arrays.sort(nums);
boolean[] visited = new boolean[nums.length];
List<Integer> permutation = new ArrayList<Integer>();
dfs(nums, visited, permutation, results);
return results;
}
private void dfs(int[] nums, boolean[] visited, List<Integer> permutation, List<List<Integer>> results) {
if (nums.length == permutation.size()) {
results.add(new ArrayList<Integer>(permutation));
return;
}
for (int i = 0; i < nums.length; i++) {
if (visited[i]) continue;
// skipping causes duplicate situations
if (i > 0 && nums[i] == nums[i - 1] && visited[i - 1] == false) {
// 假设输入的数组为[1,1,2]。则当第一个1被添加进结果集时,可以继续使用第二个1作为元素添加进结果集从而生成112。同理,当试图将第二个1作为第一个元素添加进结果集时,只要第一个1还没有被使用过,则不可以使用第二个1。因此,112不会被重复的添加进结果集。
continue;
}
permutation.add(nums[i]);
visited[i] = true;
dfs(nums, visited, permutation, results);
visited[i] = false;
permuation.remove(permutation.size() - 1);
}
}
}
Approach 2: 字典序
class Solution {
public ArrayList<ArrayList<Integer>> permuteUnique(ArrayList<Integer> nums) {
ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();
if (nums == null || nums.size() == 0) return result;
// deep copy
List<Integer> perm = new ArrayList<Integer>(nums);
Collections.sort(perm);
while (true) {
result.add(new ArrayList<Integer>(perm));
// step1: search the first num[k] < nums[k+1] backward
int k = -1;
for (int i = perm.size() - 2; i >= 0; i--) 【
if (perm.get(i) < perm.get(i + 1)) {
k = i;
break;
}
}
// if current rank is the largest, exit while loop
if (k == -1) break;
// step2: search teh first perm[k] < perm[l] backward
int l = perm.size() - 1;
while (l > k && perm.get(l) <= perm.get(k)) l--;
// step3: swap perm[k] with perm[l]
Collections.swap(perm, k, l);
// step4: reverse between k+1 and perm.length - 1
reverse(perm, k+1, perm.size() - 1);
}
return result;
}
private void reverse(List<Integer> nums, int left, int right) {
for (int l = left, r = right; l < r; l++, r--) {
Collections.swap(nums, l, r);
}
}
}
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