【力扣】17. 电话号码的字母组合

给定一个仅包含数字 2-9 的字符串，返回所有它能表示的字母组合。答案可以按 任意顺序 返回。

给出数字到字母的映射如下（与电话按键相同）。注意 1 不对应任何字母。

示例 1：

输入：digits = "23"
输出：["ad","ae","af","bd","be","bf","cd","ce","cf"]

示例 2：

输入：digits = ""
输出：[]

示例 3：

输入：digits = "2"
输出：["a","b","c"]

提示：

• 0 <= digits.length <= 4
• digits[i] 是范围 ['2', '9'] 的一个数字。

题解：

回溯法：

func letterCombinations(digits string) []string {
	letterMap := []string{"", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"}
	var s string
	var result []string
	var backtracking func(digits string, index int)
	backtracking = func(digits string, index int) {
		if index == len(digits) {
			result = append(result, s)
			return
		}
		digit := digits[index] - '0'
		letter := letterMap[digit]
		for i := 0; i < len(letter); i++ {
			s += string(letter[i])
			backtracking(digits, index+1)
			s = s[:len(s)-1]
		}
	}
	if digits != "" {
		backtracking(digits, 0)
	}
	return result
}

回溯法（隐藏了回溯过程）：

func letterCombinations(digits string) []string {
	letterMap := []string{"", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"}
	//var s string
	var result []string
	var backtracking func(digits string, index int, s string)
	backtracking = func(digits string, index int, s string) {
		if index == len(digits) {
			result = append(result, s)
			return
		}
		digit := digits[index] - '0'
		letter := letterMap[digit]
		for i := 0; i < len(letter); i++ {
			backtracking(digits, index+1, s+string(letter[i]))
		}
	}
	if digits != "" {
		backtracking(digits, 0, "")
	}
	return result
}