1048.最长字符串链

链接：1048.最长字符串链
难度：Medium
标签：数组、哈希表、双指针、字符串、动态规划、排序
简介：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 。

题解 1 - rust

• 编辑时间：2023-04-27
• 执行用时：28ms
• 内存消耗：2.1MB
• 编程语言：rust
• 解法介绍：同上。

fn cmp(s1: &[u8], s2: &[u8], i1: usize, i2: usize, err: usize) -> bool {
    if i1 == s1.len() {
        i2 + err == s2.len()
    } else if i2 == s2.len() {
        i1 + err == s1.len()
    } else if s1[i1] == s2[i2] {
        cmp(s1, s2, i1 + 1, i2 + 1, err)
    } else if err == 0 {
        false
    } else {
        cmp(s1, s2, i1 + 1, i2, err - 1) || cmp(s1, s2, i1, i2 + 1, err - 1)
    }
}
impl Solution {
    pub fn longest_str_chain(mut words: Vec<String>) -> i32 {
        words.sort_by_key(|v| v.len());
        let n = words.len();
        let mut res = 1;
        let mut dp = vec![1; n];
        for i in 0..n {
            for j in 0..i {
                if words[i].len() == words[j].len() {
                    break;
                }
                let s1 = words[i].as_bytes();
                if cmp(words[i].as_bytes(), words[j].as_bytes(), 0, 0, 1) {
                    dp[i] = dp[i].max(dp[j] + 1);
                }
            }
            res = res.max(dp[i]);
        }
        res as i32
    }
}

题解 2 - cpp

• 编辑时间：2023-04-27
• 执行用时：96ms
• 内存消耗：12.8MB
• 编程语言：cpp
• 解法介绍：遍历。

class Solution {
public:
    bool cmp(string &s1, string &s2, int i1 = 0, int i2 = 0, int err = 1) {
        if (i1 == s1.size()) return i2 + err == s2.size();
        if (i2 == s2.size()) return i1 + err == s1.size();
        if (s1[i1] == s2[i2]) return cmp(s1, s2, i1 + 1, i2 + 1, err);
        if (err == 0) return false;
        return cmp(s1, s2, i1 + 1, i2, err - 1) || cmp(s1, s2, i1, i2 + 1, err - 1);
    }
    int longestStrChain(vector<string>& words) {
        sort(words.begin(), words.end(), [&](auto &a, auto &b) { return a.size() < b.size(); });
        int n = words.size(), res = 1;
        vector<int> dp(n, 1);
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < i; j++) {
                if (words[j].size() == words[i].size()) break;
                if (cmp(words[i], words[j])) dp[i] = max(dp[i], dp[j] + 1);
            }
            res = max(res, dp[i]);
        }
        return res;
    }
};

题解 3 - python

• 编辑时间：2023-04-27
• 执行用时：1836ms
• 内存消耗：15.2MB
• 编程语言：python
• 解法介绍：同上。

def cmp(s1: str, s2: str, i1: int, i2: int, err: int):
    if i1 == len(s1):
        return i2 + err == len(s2)
    if i2 == len(s2):
        return i1 + err == len(s1)
    if s1[i1] == s2[i2]:
        return cmp(s1, s2, i1 + 1, i2 + 1, err)
    if err == 0:
        return False
    return cmp(s1, s2, i1 + 1, i2, err - 1) or cmp(s1, s2, i1, i2 + 1, err - 1)

class Solution:
    def longestStrChain(self, words: List[str]) -> int:
        words.sort(key=lambda v: len(v))
        n = len(words)
        res = 1
        dp = [1] * n
        for i in range(n):
            for j in range(i):
                if len(words[j]) == len(words[i]):
                    break
                if cmp(words[i], words[j], 0, 0, 1):
                    dp[i] = max(dp[i], dp[j] + 1)
            res = max(res, dp[i])
        return res