There are many permutation and combination problems in lintcode. If you want to write your own code for these problems, you need to use recursive thinking. .
Like array arrangement, python also has built-in library functions for permutation and combination. It is really cool to use these library functions directly in the process of writing questions.
Arranged function:
import itertools print(list(itertools.permutations([1,2,3,4],3)))
The above code directly prints the full arrangement of the 3 numbers selected in [1, 2, 3, 4]. If the second parameter is not filled, the default is the length of the array.
Combined functions:
import itertools print(list(itertools.combinations([1,2,3,4],3)))
Similar to the situation above.
Take the specific topic of lintcode as an example
full array
Given a list of numbers, return all possible permutations of it.
Precautions
You can assume there are no repeating numbers.
import itertools
class Solution:
"""
@param nums: A list of Integers.
@return: A list of permutations.
"""
def permute(self, nums):
# write your code here
result=[]
permutation_result=list(itertools.permutations(nums))
for i in permutation_result:
result.append(list(i))
return result
If written in a non-recursive form, it would look like this:
class Solution:
"""
@param nums: A list of Integers.
@return: A list of permutations.
"""
def permute(self, nums):
if nums is None:
return []
if nums == []:
return [[]]
nums = sorted(nums)
permutation = []
stack = [-1]
permutations = []
while len(stack):
index = stack.pop()
index += 1
while index < len(nums):
if nums[index] not in permutation:
break
index += 1
else:
if len(permutation):
permutation.pop()
continue
stack.append(index)
stack.append(-1)
permutation.append(nums[index])
if len(permutation) == len(nums):
permutations.append(list(permutation))
return permutations
kth permutation
Given n and k , find 123..nthe kth permutation of the permutations formed.
Precautions
1 ≤ n ≤ 9
For n = 3, all permutations are as follows:
123
132
213
231
312
321
如果 k = 4, 第4个排列为，231.
import itertools
class Solution:
"""
@param n: n
@param k: the k-th permutation
@return: a string, the k-th permutation
"""
def getPermutation(self, n, k):
result = list(itertools.permutations(range(1,n+1)))[k-1]
string=""
for i in result:
string +=str(i)
return string
k数和 II
给定n个不同的正整数，整数k（1<= k <= n）以及一个目标数字。
在这n个数里面找出K个数，使得这K个数的和等于目标数字，你需要找出所有满足要求的方案。
给出[1,2,3,4]，k=2， target=5，返回 [[1,4],[2,3]]
从n个数找出k个数，是组合问题，得到组合后，一个一个和是否为目标数字，即得到结果。
import itertools
class Solution:
"""
@param A: An integer array.
@param k: A positive integer (k <= length(A))
@param target: Integer
@return a list of lists of integer
"""
def kSumII(self, A, k, target):
# write your code here
result=list(itertools.combinations(A,k))
lists=[]
for i in result:
if sum(i)==target:
lists.append(list(i))
return (lists)
三数之和
给出一个有n个整数的数组S，在S中找到三个整数a, b, c，找到所有使得a + b + c = 0的三元组。
注意事项
在三元组(a, b, c)，要求a <= b <= c。
结果不能包含重复的三元组。
如S = {-1 0 1 2 -1 -4}, 你需要返回的三元组集合的是：
(-1, 0, 1)
(-1, -1, 2)
import itertools
class Solution:
"""
@param numbersbers : Give an array numbersbers of n integer
@return : Find all unique triplets in the array which gives the sum of zero.
"""
def threeSum(self, numbers):
"""
首先要对numbers进行清洗，如果numbers中有的数出现了超过两次，则让其最多出现两次。
例如numbers由100个1和100个-1组成。清洗后的数组的结果和未清洗的结果都是相同的。
但0就不一样了，因为3个0的和也是0，所以0最多可以出现3次
清洗完后，对numbers进行组合，判断组合是否和为0。
注意，返回的结果要按数字大小排序，而且不能有重复的结果。
"""
CleanNumbers=[]
setNumbers=set(numbers)
for i in setNumbers:
if numbers.count(i)>=2:
CleanNumbers.append(i)
CleanNumbers.append(i)
else:
CleanNumbers.append(i)
if numbers.count(0)>=3:
CleanNumbers.append(0)
result=[]
combinations=list(itertools.combinations(CleanNumbers,3))
for i in combinations:
if sum(i)==0 and sorted(list(i)) not in result:
i=list(i)
i.sort()
result.append(i)
return sorted(result)
四数之和
给一个包含n个数的整数数组S，在S中找到所有使得和为给定整数target的四元组(a, b, c, d)。
注意事项
四元组(a, b, c, d)中，需要满足a <= b <= c <= d
答案中不可以包含重复的四元组。
例如，对于给定的整数数组S=[1, 0, -1, 0, -2, 2] 和 target=0. 满足要求的四元组集合为：
(-1, 0, 0, 1)
(-2, -1, 1, 2)
(-2, 0, 0, 2)
这道题和上面的三数之和是一模一样的思路，代码如下：
import itertools
class Solution:
"""
@param numbersbers : Give an array numbersbersbers of n integer
@param target : you need to find four elements that's sum of target
@return : Find all unique quadruplets in the array which gives the sum of
zero.
"""
def fourSum(self, numbers, target):
# write your code here
setNumbers=set(numbers)
cleanedNumbers=[]
for i in setNumbers:
if numbers.count(i)>=3:
cleanedNumbers.append(i)
cleanedNumbers.append(i)
cleanedNumbers.append(i)
elif numbers.count(i)==2:
cleanedNumbers.append(i)
cleanedNumbers.append(i)
else:
cleanedNumbers.append(i)
if numbers.count(0)>=4:
cleanedNumbers.append(0)
result = []
combinations = list(itertools.combinations(cleanedNumbers, 4))
for i in combinations:
if sum(i) == target and sorted(list(i)) not in result:
result.append(sorted(list(i)))
return sorted(result)
最接近的三数之和
给一个包含 n 个整数的数组 S, 找到和与给定整数 target 最接近的三元组，返回这三个数的和。
注意事项
只需要返回三元组之和，无需返回三元组本身
例如 S = [-1, 2, 1, -4] and target = 1. 和最接近 1 的三元组是 -1 + 2 + 1 = 2.
import itertools
class Solution:
"""
@param numbers: Give an array numbers of n integer
@param target : An integer
@return : return the sum of the three integers, the sum closest target.
"""
def threeSumClosest(self, numbers, target):
# write your code here
cleanedNumbers=[]
setNumbers=set(numbers)
for i in setNumbers:
if numbers.count(i)>=3:
cleanedNumbers.append(i)
cleanedNumbers.append(i)
cleanedNumbers.append(i)
elif numbers.count(i)==2:
cleanedNumbers.append(i)
cleanedNumbers.append(i)
else:
cleanedNumbers.append(i)
combinations = list(itertools.combinations(cleanedNumbers,3))
minNumber=sum(combinations[0])
for i in combinations:
if abs(sum(i)-target)<abs(minNumber-target):
minNumber=sum(i)
return minNumber
带重复元素的排列
给出一个具有重复数字的列表，找出列表所有不同的排列。
给出列表 [1,2,2]，不同的排列有：
[
[1,2,2],
[2,1,2],
[2,2,1]
]import itertools
class Solution:
"""
@param nums: A list of integers.
@return: A list of unique permutations.
"""
def permuteUnique(self, nums):
# write your code here
result=[]
permutations=list(itertools.permutations(nums))
for i in permutations:
if list(i) not in result:
result.append(list(i))
return result