题目描述
给出一个具有重复数字的列表,找出列表所有不同的排列。
样例1
输入:[1,1]
输出:
[
[1,1]
]
样例2
输入:[1,2,2]
输出:
[
[1,2,2],
[2,1,2],
[2,2,1]
]
解题思路
使用排列式深度优先搜索算法。
和没有重复元素的 Permutation 一题相比,只加了两句话:
Arrays.sort(nums) // 排序这样所有重复的数
if (i > 0 && nums[i] == nums[i - 1] && !visited[i - 1]) { continue; } // 跳过会造成重复的情况
java题解
public class Solution {
/*
* @param : A list of integers
* @return: A list of unique permutations
*/
public List<List<Integer>> permuteUnique(int[] nums) {
List<List<Integer>> results = new ArrayList<>();
if (nums == null) {
return results;
}
Arrays.sort(nums);
dfs(nums, new boolean[nums.length], new ArrayList<Integer>(), 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;
}
if (i > 0 && nums[i] == nums[i - 1] && !visited[i - 1]) {
continue;
}
permutation.add(nums[i]);
visited[i] = true;
dfs(nums, visited, permutation, results);
visited[i] = false;
permutation.remove(permutation.size() - 1);
}
}
};
C++题解
class Solution {
private:
void helper(vector<vector<int> > &results,
vector<int> &permutation,
vector<int> &nums,
bool used[]) {
if (nums.size() == permutation.size()) {
results.push_back(permutation);
return;
}
for (int i = 0; i < nums.size(); i++) {
if (used[i]) {
continue;
}
if (i > 0 && used[i - 1] == false && nums[i] == nums[i-1]) {
continue;
}
used[i] = true;
permutation.push_back(nums[i]);
helper(results, permutation, nums, used);
permutation.pop_back();
used[i] = false;
}
}
public:
vector<vector<int> > permuteUnique(vector<int> &nums) {
vector<vector<int> > results;
vector<int> permutation;
bool used[nums.size()];
for (int i = 0; i < nums.size(); i++) {
used[i] = false;
}
sort(nums.begin(), nums.end());
helper(results, permutation, nums, used);
return results;
}
};
python题解
class Solution:
"""
@param nums: A list of integers.
@return: A list of unique permutations.
"""
def permuteUnique(self, nums):
# write your code here
def _permute(result, temp, nums):
if nums == []:
result += [temp]
else:
for i in range(len(nums)):
if i > 0 and nums[i] == nums[i-1]:
continue
_permute(result, temp + [nums[i]], nums[:i] + nums[i+1:])
if nums is None:
return []
if len(nums) == 0:
return [[]]
result = []
_permute(result, [], sorted(nums))
return result