Question

1048. 最长字符串链

English Version

题目描述

给出一个单词数组 words ，其中每个单词都由小写英文字母组成。

如果我们可以 不改变其他字符的顺序 ，在 wordA 的任何地方添加 恰好一个 字母使其变成 wordB ，那么我们认为 wordA 是 wordB 的 前身 。

• 例如，"abc" 是 "abac" 的 前身 ，而 "cba" 不是 "bcad" 的 前身

词链是单词 [word_1, word_2, ..., word_k] 组成的序列，k >= 1，其中 word1 是 word2 的前身，word2 是 word3 的前身，依此类推。一个单词通常是 k == 1 的 单词链 。

从给定单词列表 words 中选择单词组成词链，返回 词链的 最长可能长度 。
 

示例 1：

输入：words = ["a","b","ba","bca","bda","bdca"]
输出：4
解释：最长单词链之一为 ["a","ba","bda","bdca"]

示例 2:

输入：words = ["xbc","pcxbcf","xb","cxbc","pcxbc"]
输出：5
解释：所有的单词都可以放入单词链 ["xb", "xbc", "cxbc", "pcxbc", "pcxbcf"].

示例 3:

输入：words = ["abcd","dbqca"]
输出：1
解释：字链["abcd"]是最长的字链之一。
["abcd"，"dbqca"]不是一个有效的单词链，因为字母的顺序被改变了。

 

提示：

• 1 <= words.length <= 1000
• 1 <= words[i].length <= 16
• words[i] 仅由小写英文字母组成。

解法

方法一

class Solution:
  def longestStrChain(self, words: List[str]) -> int:
    def check(w1, w2):
      if len(w2) - len(w1) != 1:
        return False
      i = j = cnt = 0
      while i < len(w1) and j < len(w2):
        if w1[i] != w2[j]:
          cnt += 1
        else:
          i += 1
        j += 1
      return cnt < 2 and i == len(w1)

    n = len(words)
    dp = [1] * (n + 1)
    words.sort(key=lambda x: len(x))
    res = 1
    for i in range(1, n):
      for j in range(i):
        if check(words[j], words[i]):
          dp[i] = max(dp[i], dp[j] + 1)
      res = max(res, dp[i])
    return res

class Solution {
  public int longestStrChain(String[] words) {
    Arrays.sort(words, Comparator.comparingInt(String::length));
    int res = 0;
    Map<String, Integer> map = new HashMap<>();
    for (String word : words) {
      int x = 1;
      for (int i = 0; i < word.length(); ++i) {
        String pre = word.substring(0, i) + word.substring(i + 1);
        x = Math.max(x, map.getOrDefault(pre, 0) + 1);
      }
      map.put(word, x);
      res = Math.max(res, x);
    }
    return res;
  }
}

class Solution {
public:
  int longestStrChain(vector<string>& words) {
    sort(words.begin(), words.end(), [&](string a, string b) { return a.size() < b.size(); });
    int res = 0;
    unordered_map<string, int> map;
    for (auto word : words) {
      int x = 1;
      for (int i = 0; i < word.size(); ++i) {
        string pre = word.substr(0, i) + word.substr(i + 1);
        x = max(x, map[pre] + 1);
      }
      map[word] = x;
      res = max(res, x);
    }
    return res;
  }
};

func longestStrChain(words []string) int {
  sort.Slice(words, func(i, j int) bool { return len(words[i]) < len(words[j]) })
  res := 0
  mp := make(map[string]int)
  for _, word := range words {
    x := 1
    for i := 0; i < len(word); i++ {
      pre := word[0:i] + word[i+1:len(word)]
      x = max(x, mp[pre]+1)
    }
    mp[word] = x
    res = max(res, x)
  }
  return res
}

function longestStrChain(words: string[]): number {
  words.sort((a, b) => a.length - b.length);
  let ans = 0;
  let hashTable = new Map();
  for (let word of words) {
    let c = 1;
    for (let i = 0; i < word.length; i++) {
      let pre = word.substring(0, i) + word.substring(i + 1);
      c = Math.max(c, (hashTable.get(pre) || 0) + 1);
    }
    hashTable.set(word, c);
    ans = Math.max(ans, c);
  }
  return ans;
}

use std::collections::HashMap;

impl Solution {
  #[allow(dead_code)]
  pub fn longest_str_chain(words: Vec<String>) -> i32 {
    let mut words = words;
    let mut ret = 0;
    let mut map: HashMap<String, i32> = HashMap::new();

    // Sort the words vector first
    words.sort_by(|lhs, rhs| { lhs.len().cmp(&rhs.len()) });

    // Begin the "dp" process
    for w in words.iter() {
      let n = w.len();
      let mut x = 1;

      for i in 0..n {
        let s = w[..i].to_string() + &w[i + 1..];
        let v = map.entry(s.clone()).or_default();
        x = std::cmp::max(x, *v + 1);
      }

      map.insert(w.clone(), x);

      ret = std::cmp::max(ret, x);
    }

    ret
  }
}

方法二

class Solution:
  def longestStrChain(self, words: List[str]) -> int:
    words.sort(key=lambda x: len(x))
    res = 0
    mp = {}
    for word in words:
      x = 1
      for i in range(len(word)):
        pre = word[:i] + word[i + 1 :]
        x = max(x, mp.get(pre, 0) + 1)
      mp[word] = x
      res = max(res, x)
    return res