Find all valid combinations of?k
?numbers that sum up to?n
?such that the following conditions are true:
1
?through?9
?are used.Return?a list of all possible valid combinations. The list must not contain the same combination twice, and the combinations may be returned in any order.
Example 1:
Input: k = 3, n = 7 Output: [[1,2,4]] Explanation: 1 + 2 + 4 = 7 There are no other valid combinations.
思路:和昨天的那题很相似,不过结束条件不太一样。以及在回溯的时候需要多记录一个值。
class Solution(object):
def backtrack(self, target, k, currentSum, startIndex, path, result):
if currentSum > target: # early stop
return
if len(path) == k: # 判断是否满足条件
if currentSum == target:
result.append(path[:])
return
for i in range(startIndex, 9-(k-len(path)) + 2): # 如果已经有两个元素,而总共需要5个,那么9-(5-2)+ 2 = 8,最后一个值取不到,所以有7个值可取。
currentSum += i
path.append(i)
self.backtrack(target, k, currentSum, i+1, path, result) # 注意这里是i不是startIndex
path.pop()
currentSum -= i
def combinationSum3(self, k, n):
"""
:type k: int
:type n: int
:rtype: List[List[int]]
"""
result = []
self.backtrack(n,k,0,1,[],result)
return result
这里一个数字对应三个字母,所以相当于二叉树变得更宽了(每一层有更多的子节点)。另外注意,这里用来记录每条“路径”的是string,所以不能用pop()来回溯,而要用[:-1]
class Solution(object):
def __init__(self):
self.letterMap = [
"", #0
"", #
"abc",
"def",
"ghi",
"jkl", # 5
"mno", # 6
"pqrs", # 7
"tuv", # 8
"wxyz" # 9
]
self.result = []
self.s = ""
def backtrack(self, digits, index):
if index == len(digits):
self.result.append(self.s)
return
digit = int(digits[index])
letters = self.letterMap[digit]
for i in range(len(letters)):
self.s += letters[i]
self.backtrack(digits, index+1)
self.s = self.s[:-1] # this is a string, so the pop() cannot be used
def letterCombinations(self, digits):
"""
:type digits: str
:rtype: List[str]
"""
if len(digits) == 0:
return self.result
self.backtrack(digits, 0)
return self.result