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字母大小写全排列-784 #106

@sl1673495

Description

@sl1673495

给定一个字符串 S,通过将字符串 S 中的每个字母转变大小写,我们可以获得一个新的字符串。返回所有可能得到的字符串集合。

示例:
输入: S = "a1b2"
输出: ["a1b2", "a1B2", "A1b2", "A1B2"]

输入: S = "3z4"
输出: ["3z4", "3Z4"]

输入: S = "12345"
输出: ["12345"]
注意:

S 的长度不超过12。
S 仅由数字和字母组成。

来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/letter-case-permutation
著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。

思路

这题不需要交换顺序排列组合,所以会简单点,只需要把当前第一位的字母分别大小写处理(如果是数字就直接不动)拼接在前面已经处理好的字符串后,带上截取掉第一位后的剩余字符串交给下一轮递归,直到拼接的字符长度符合 S 的长度即可终止。

/**
 * @param {string} S
 * @return {string[]}
 */
let letterCasePermutation = function (S) {
  let res = []

  let helper = (prev, rest) => {
    if (prev.length === S.length) {
      res.push(prev)
      return
    }

    let char = rest[0]
    let word1 = prev + char
    let nextRest = rest.substring(1)

    if (!isNaN(Number(char))) {
      helper(word1, nextRest)
      return
    } else {
      let upperChar = char.toUpperCase()
      let char2 = upperChar === char ? char.toLowerCase() : upperChar
      let word2 = prev + char2

      helper(word1, nextRest)
      helper(word2, nextRest)
    }
  }

  helper("", S)

  return res
}

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