电话号码的字母组合

LeetCode题目链接：https://leetcode.cn/problems/letter-combinations-of-a-phone-number/

难度：中等

分析

本题属于典型的递归回溯的题型，根据数字查找可使用的字母，在进行遍历递归回溯组合出所有可能的组合即可

1688805900848

思路

  1. 递归纵向遍历数字串（递归过程中当前遍历的索引即可实现递归纵向遍历）
  2. 根据数字取出对应的字母集合
  3. 遍历该集合组合结果

递归结束条件：当结果集中的字母和所给数字一样多时，即遍历结束

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const (
    CHAR_ONE   = '1'
    CHAR_TWO   = '2'
    CHAR_THREE = '3'
    CHAR_FOUR  = '4'
    CHAR_FIVE  = '5'
    CHAR_SIX   = '6'
    CHAR_SEVEN = '7'
    CHAR_EIGHT = '8'
    CHAR_NINE  = '9'
)

var (
    ONE_LETTERS   = []uint8{}
    TWO_LETTERS   = []uint8{'a', 'b', 'c'}
    THREE_LETTERS = []uint8{'d', 'e', 'f'}
    FOUR_LETTERS  = []uint8{'g', 'h', 'i'}
    FIVE_LETTERS  = []uint8{'j', 'k', 'l'}
    SIX_LETTERS   = []uint8{'m', 'n', 'o'}
    SEVEN_LETTERS = []uint8{'p', 'q', 'r', 's'}
    EIGHT_LETTERS = []uint8{'t', 'u', 'v'}
    NINE_LETTERS  = []uint8{'w', 'x', 'y', 'z'}
)

var NUM_LETTERS_COR_MAP = map[uint8][]uint8{
    CHAR_ONE:   ONE_LETTERS,
    CHAR_TWO:   TWO_LETTERS,
    CHAR_THREE: THREE_LETTERS,
    CHAR_FOUR:  FOUR_LETTERS,
    CHAR_FIVE:  FIVE_LETTERS,
    CHAR_SIX:   SIX_LETTERS,
    CHAR_SEVEN: SEVEN_LETTERS,
    CHAR_EIGHT: EIGHT_LETTERS,
    CHAR_NINE:  NINE_LETTERS,
}

func recurve(digits string, ret *[]string, index int, cur []uint8) {
    if index == len(digits){
        // 递归结束，加入结果集
        if len(cur) > 0 {
            *ret = append(*ret, string(cur))
        }
        return
    }

    digit := digits[index]
    letters := NUM_LETTERS_COR_MAP[digit]
    for _, letter := range letters {
        recurve(digits, ret, index+1, append(cur, letter))
    }
}

func letterCombinations(digits string) []string {
    var ret []string
    recurve(digits, &ret, 0, []uint8{})
    return ret
}