17. 电话号码的字母组合 @Python @Go

原题

https://leetcode.cn/problems/letter-combinations-of-a-phone-number/description/

思路

回溯法

复杂度

时间：O(n * 4 ** (n))
空间：O(n)

Python代码

class Solution:
    def letterCombinations(self, digits: str) -> List[str]:
        map = dict(
            zip("23456789", ["abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"])
        )
        ans = []
                
        def backtrack(i, path):
            if i == len(digits):
                ans.append(path)
                return 
            for char in map[digits[i]]:
                backtrack(i+1, path + char)
        
        if len(digits) == 0:
            return ans
        backtrack(0, '')
        return ans
         

Go代码

func letterCombinations(digits string) []string {
	var ans []string
	if len(digits) == 0 {
		return ans
	}
	// 创建map
	m := map[byte]string{}
	letters := "23456789"
	coms := []string{"abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"}
	for i, ch := range letters {
		m[byte(ch)] = coms[i]
	}
	// 匿名函数
	var backtrack func(int, string)
	backtrack = func(i int, path string) {
		if i == len(digits) {
			ans = append(ans, path)
			return
		}
		for _, char := range m[byte(digits[i])] {
			backtrack(i+1, path+string(char))
		}
	}
    // 回溯法
	backtrack(0, "")
	return ans
}