Encode String with Shortest Length

最新推荐文章于 2023-12-21 05:45:00 发布

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最新推荐文章于 2023-12-21 05:45:00 发布

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分类专栏： Dynamic Programming

本文链接：https://blog.youkuaiyun.com/u013325815/article/details/106914037

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本文深入探讨了一种字符串压缩算法，该算法通过动态规划方法寻找最短的编码方式，特别适用于包含重复子串的输入。文章提供了详细的实现思路，包括如何通过检查子串的重复模式来实现有效的压缩。

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Given a non-empty string, encode the string such that its encoded length is the shortest.

The encoding rule is: k[encoded_string], where the encoded_string inside the square brackets is being repeated exactly k times.

Note:

k will be a positive integer and encoded string will not be empty or have extra space.
You may assume that the input string contains only lowercase English letters. The string's length is at most 160.
If an encoding process does not make the string shorter, then do not encode it. If there are several solutions, return any of them is fine.

Example 1:

Input: "aaa"
Output: "aaa"
Explanation: There is no way to encode it such that it is shorter than the input string, so we do not encode it.

Example 2:

Input: "aaaaa"
Output: "5[a]"
Explanation: "5[a]" is shorter than "aaaaa" by 1 character.

Example 3:

Input: "aaaaaaaaaa"
Output: "10[a]"
Explanation: "a9[a]" or "9[a]a" are also valid solutions, both of them have the same length = 5, which is the same as "10[a]".

Example 4:

Input: "aabcaabcd"
Output: "2[aabc]d"
Explanation: "aabc" occurs twice, so one answer can be "2[aabc]d".

Example 5:

Input: "abbbabbbcabbbabbbc"
Output: "2[2[abbb]c]"
Explanation: "abbbabbbc" occurs twice, but "abbbabbbc" can also be encoded to "2[abbb]c", so one answer can be "2[2[abbb]c]".

思路：这题因为中间任意重复字符都可以形成重复，所以这是一个长度区间型动态规划，从len最小的开始计算，一直计算到最大的。dp[i][j]代表从i,....j 最小的compress string，dp[i][j] , 可以从两个方面得到, dp[i][k] + dp[k + 1][j] 如果合起来比当前小，update

如果dp[i][j] 自己能够组成最小的compression，也需要update。注意判断最小的compress string的时候，用pattern去判断，

pattern != null && substr.length() % pattern.length() == 0 && substr.replaceAll(pattern, "").length() == 0) 这句代码很巧妙，学习了。另外，长度dp的模板要记住：

for(int len = 1; len <= n; len++) {
for(int i = 0; i + len - 1 < n; i++) {
int j = i + len - 1;

class Solution {
    public String encode(String s) {
        int n = s.length();
        String[][] dp = new String[n][n];
        
        for(int len = 1; len <= n; len++) {
            for(int i = 0; i + len - 1 < n; i++) {
                int j = i + len - 1;
                String substr = s.substring(i, j + 1);
                dp[i][j] = substr;
                // 看左右两边能否组合成最小的string；
                if(j - i >= 4) {
                    for(int k = i; k < j; k++) {
                        if(dp[i][k].length() + dp[k + 1][j].length() < dp[i][j].length()) {
                            dp[i][j] = dp[i][k] + dp[k + 1][j];
                        }
                    }
                }
                
                // 看自己能否组成最小的string；
                for(int k = 0; k < substr.length(); k++) {
                    // 遍历所有的可能pattern；
                    String pattern = substr.substring(0, k + 1);
                    if(pattern != null
                      && (substr.length() % pattern.length() == 0)
                      && substr.replaceAll(pattern, "").length() == 0) {
                        String candidate = substr.length() / pattern.length() 
                            + "[" + dp[i][i + k] + "]";
                        if(candidate.length() < dp[i][j].length()) {
                            dp[i][j] = candidate;
                        }
                    }
                }
            }
        }
        return dp[0][n - 1];
    }
}