算法练习Day21 （Leetcode/Python-回溯算法）

216. Combination Sum III

Find all valid combinations of k numbers that sum up to n such that the following conditions are true:

• Only numbers 1 through 9 are used.
• Each number is used at most once.

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 

17. Letter Combinations of a Phone Number

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