90.Decode Ways

A message containing letters from A-Z can be encoded into numbers using the following mapping:

'A' -> "1"
'B' -> "2"
...
'Z' -> "26"

To decode an encoded message, all the digits must be grouped then mapped back into letters using the reverse of the mapping above (there may be multiple ways). For example, "11106" can be mapped into:

• "AAJF" with the grouping (1 1 10 6)
• "KJF" with the grouping (11 10 6)

Note that the grouping (1 11 06) is invalid because "06" cannot be mapped into 'F' since "6" is different from "06".

Given a string s containing only digits, return the number of ways to decode it.

The answer is guaranteed to fit in a 32-bit integer.

Example 1:

Input: s = "12"
Output: 2
Explanation: "12" could be decoded as "AB" (1 2) or "L" (12).

Example 2:

Input: s = "226"
Output: 3
Explanation: "226" could be decoded as "BZ" (2 26), "VF" (22 6), or "BBF" (2 2 6).

Example 3:

Input: s = "0"
Output: 0
Explanation: There is no character that is mapped to a number starting with 0.
The only valid mappings with 0 are 'J' -> "10" and 'T' -> "20", neither of which start with 0.
Hence, there are no valid ways to decode this since all digits need to be mapped.

Example 4:

Input: s = "06"
Output: 0
Explanation: "06" cannot be mapped to "F" because of the leading zero ("6" is different from "06").

Constraints:

• 1 <= s.length <= 100
• s contains only digits and may contain leading zero(s).

Analyze

该题拆分子问题较为不易, 题目思路可以转化为:

• 下标为 start 的拆解之和等于
  • 下标为 start + 1 的拆解之和
  • 下标为 start + 2 的拆解之和(前提: [start, start + 2) 值合理)

解法一: 记忆化递归(可二刷)

/**
 * @param {string} s
 * @return {number}
 */
var numDecodings = function(s) {
  const cache = {}
  return judge(s, 0, cache)
}

var judge = function(s, start, cache) {
  const length = s.length
  if (s[start] === '0') return 0
  if (start === s.length) {
    return 1
  }
  if (cache[start]) return cache[start]

  let res = judge(s, start + 1, cache)
  if (isValidValue(s.slice(start, start + 2))) {
    res += judge(s, start + 2, cache)
  }
  return cache[start] = res
}

var isValidValue = function(value) {
  if (!value || value.length < 2) return false
  const valueToNum = Number(value)
  return valueToNum <= 26
}

解法二: 动态规划

cache[i] 表示第 i 位上有几种拆解方法。

/**
 * @param {string} s
 * @return {number}
 */
var numDecodings = function(s) {
  if (s[0] === '0') {
    return 0
  }

  const cache = {
    [`${s.length}`]: 1
  }

  for (let i = s.length - 1; i >= 0; i--) {
    if (s.slice(i, i + 1) === '0') {
      cache[i] = 0
    } else {
      cache[i] = cache[i + 1]

      if (isValidValue(s.slice(i, i + 2))) {
        cache[i] = cache[i] + cache[i + 2]
      }
    }
  }

  return cache['0']
}

var isValidValue = function(value) {
  if (!value || value.length < 2) return false
  const valueToNum = Number(value)
  return valueToNum <= 26
}