算法训练营

216.组合总和III

题目链接: 216.组合总和III
文档讲解： 代码随想录
题目难度：中等

思路:递归法
时间复杂度：O(n*2^n)；空间复杂度：O(n)
下面展示 代码：

class Solution:
    def backtracking(self, n,k,start,sum, path,res):
        if sum > n:
            return
        if len(path) == k and sum == n:
            res.append(path[:])  # 终止条件
            return
        else:
            for i in range(start, 9-(k-len(path)) + 2):
                path.append(i)
                sum += i
                self.backtracking(n,k,i + 1,sum, path,res)
                path.pop() # 回溯
                sum -= i
    def combinationSum3(self, k: int, n: int) -> List[List[int]]:
        # 和为n的k个数,(1-9)
        if k == 1:
            if 1 <= n <= 9:
                return [n]
            else:
                return []
        else:
            res = []
            self.backtracking(n, k, 1,0,[], res)
            return res

17.电话号码的字母组合

题目链接: 17.电话号码的字母组合
文档讲解： 代码随想录
题目难度：中等
思路:递归+构造号码字典
时间复杂度：O(n*2^n)；空间复杂度：O(n)
代码如下

class Solution:
    def __init__(self):
            self.letterMap = [
                "",     # 0
                "",     # 1
                "abc",  # 2
                "def",  # 3
                "ghi",  # 4
                "jkl",  # 5
                "mno",  # 6
                "pqrs", # 7
                "tuv",  # 8
                "wxyz"  # 9
            ]
            self.res = []
            self.s = ""
    def backtracking(self, digits, index):
          if len(digits) == index:
                self.res.append(self.s)
                return
          else:
                num = int(digits[index])
                string = self.letterMap[num]
                for i in range(len(string)):
                      self.s += string[i]
                      self.backtracking(digits, index + 1)
                      self.s = self.s[:-1]
                
    def letterCombinations(self, digits: str) -> List[str]:
          if len(digits) == 0:
                return self.res
          else:
                self.backtracking(digits, 0)
                return self.res