题目链接 : https://leetcode-cn.com/problems/permutations/
题目描述:
给定一个没有重复数字的序列，返回其所有可能的全排列。
示例:
输入: [1,2,3]
输出:
[
[1,2,3],
[1,3,2],
[2,1,3],
[2,3,1],
[3,1,2],
[3,2,1]
]思路:
思路1 : 库函数
Python3 itertools 文档
一行代码:
def permute(self, nums: List[int]) -> List[List[int]]:
return list(itertools.permutations(nums))思路2:
回溯算法
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代码:
python
class Solution:
def permute(self, nums: List[int]) -> List[List[int]]:
res = []
def backtrack(nums, tmp):
if not nums:
res.append(tmp)
return
for i in range(len(nums)):
backtrack(nums[:i] + nums[i+1:], tmp + [nums[i]])
backtrack(nums, [])
return resjava
class Solution {
public List<List<Integer>> permute(int[] nums) {
List<List<Integer>> res = new ArrayList<>();
int[] visited = new int[nums.length];
backtrack(res, nums, new ArrayList<Integer>(), visited);
return res;
}
private void backtrack(List<List<Integer>> res, int[] nums, ArrayList<Integer> tmp, int[] visited) {
if (tmp.size() == nums.length) {
res.add(new ArrayList<>(tmp));
return;
}
for (int i = 0; i < nums.length; i++) {
if (visited[i] == 1) continue;
visited[i] = 1;
tmp.add(nums[i]);
backtrack(res, nums, tmp, visited);
visited[i] = 0;
tmp.remove(tmp.size() - 1);
}
}
}类似题目还有:
这类题目都是同一类型的,用回溯算法!
其实回溯算法关键在于:不合适就退回上一步
然后通过约束条件, 减少时间复杂度.
大家可以从下面的解法找出一点感觉!
class Solution:
def subsets(self, nums):
if not nums:
return []
res = []
n = len(nums)
def helper(idx, temp_list):
res.append(temp_list)
for i in range(idx, n):
helper(i + 1, temp_list + [nums[i]])
helper(0, [])
return resclass Solution(object):
def subsetsWithDup(self, nums):
"""
:type nums: List[int]
:rtype: List[List[int]]
"""
if not nums:
return []
n = len(nums)
res = []
nums.sort()
# 思路1
def helper1(idx, n, temp_list):
if temp_list not in res:
res.append(temp_list)
for i in range(idx, n):
helper1(i + 1, n, temp_list + [nums[i]])
# 思路2
def helper2(idx, n, temp_list):
res.append(temp_list)
for i in range(idx, n):
if i > idx and nums[i] == nums[i - 1]:
continue
helper2(i + 1, n, temp_list + [nums[i]])
helper2(0, n, [])
return resclass Solution(object):
def permute(self, nums):
"""
:type nums: List[int]
:rtype: List[List[int]]
"""
if not nums:
return
res = []
n = len(nums)
visited = [0] * n
def helper1(temp_list,length):
if length == n:
res.append(temp_list)
for i in range(n):
if visited[i] :
continue
visited[i] = 1
helper1(temp_list+[nums[i]],length+1)
visited[i] = 0
def helper2(nums,temp_list,length):
if length == n:
res.append(temp_list)
for i in range(len(nums)):
helper2(nums[:i]+nums[i+1:],temp_list+[nums[i]],length+1)
helper1([],0)
return resclass Solution(object):
def permuteUnique(self, nums):
"""
:type nums: List[int]
:rtype: List[List[int]]
"""
if not nums:
return []
nums.sort()
n = len(nums)
visited = [0] * n
res = []
def helper1(temp_list, length):
# if length == n and temp_list not in res:
# res.append(temp_list)
if length == n:
res.append(temp_list)
for i in range(n):
if visited[i] or (i > 0 and nums[i] == nums[i - 1] and not visited[i - 1]):
continue
visited[i] = 1
helper1(temp_list + [nums[i]], length + 1)
visited[i] = 0
def helper2(nums, temp_list, length):
if length == n and temp_list not in res:
res.append(temp_list)
for i in range(len(nums)):
helper2(nums[:i] + nums[i + 1:], temp_list + [nums[i]], length + 1)
helper1([],0)
# helper2(nums, [], 0)
return resclass Solution(object):
def combinationSum(self, candidates, target):
"""
:type candidates: List[int]
:type target: int
:rtype: List[List[int]]
"""
if not candidates:
return []
if min(candidates) > target:
return []
candidates.sort()
res = []
def helper(candidates, target, temp_list):
if target == 0:
res.append(temp_list)
if target < 0:
return
for i in range(len(candidates)):
if candidates[i] > target:
break
helper(candidates[i:], target - candidates[i], temp_list + [candidates[i]])
helper(candidates,target,[])
return resclass Solution:
def combinationSum2(self, candidates: List[int], target: int) -> List[List[int]]:
if not candidates:
return []
candidates.sort()
n = len(candidates)
res = []
def backtrack(i, tmp_sum, tmp_list):
if tmp_sum == target:
res.append(tmp_list)
return
for j in range(i, n):
if tmp_sum + candidates[j] > target : break
if j > i and candidates[j] == candidates[j-1]:continue
backtrack(j + 1, tmp_sum + candidates[j], tmp_list + [candidates[j]])
backtrack(0, 0, [])
return res