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]".
这道题让我们压缩字符串，把相同的字符串用中括号括起来，然后在前面加上出现的次数，感觉还是一道相当有难度的题呢。参考了网上大神的帖子才弄懂该怎么做，这道题还是应该用DP来做。我们建立一个二维的DP数组，其中dp[i][j]表示s在[i, j]范围内的字符串的缩写形式(如果缩写形式长度大于子字符串，那么还是保留子字符串)，那么如果s字符串的长度是n，最终我们需要的结果就保存在dp[0][n-1]中，然后我们需要遍历s的所有子字符串，对于任意一段子字符串[i, j]，我们\\我们以中间任意位置k来拆分成两段，比较dp[i][k]加上dp[k+1][j]的总长度和dp[i][j]的长度，将长度较小的字符串赋给dp[i][j]，然后我们要做的就是在s中取出[i, j]范围内的子字符串t进行合并。合并的方法是我们在取出的字符串t后面再加上一个t，然后在这里面寻找子字符串t的第二个起始位置，如果第二个起始位置小于t的长度的话，说明t包含重复字符串，举个例子吧，比如 t = “abab”, 那么t+t = “abababab”，我们在里面找第二个t出现的位置为2，小于t的长度4，说明t中有重复出现，重复的个数为t.size()/pos = 2个，那么我们就要把重复的地方放入中括号中，注意中括号里不能直接放这个子字符串，而是应该从dp中取出对应位置的字符串，因为重复的部分有可能已经写成缩写形式了，比如题目中的例子5。再看一个例子，如果t = “abc”，那么t+t = “abcabc”，我们在里面找第二个t出现的位置为3，等于t的长度3，说明t中没有重复出现，那么replace就还是t。然后我们比较我们得到的replace和dp[i][j]中的字符串长度，把长度较小的赋给dp[i][j]即可，时间复杂度为O(n
3)，空间复杂度为O(n
2)，参见代码如下： 解法一：
class Solution { public: string encode(string s) { int n = s.size(); vector<vector<string>> dp(n, vector<string>(n, "")); for (int step = 1; step <= n; ++step) { for (int i = 0; i + step - 1 < n; ++i) { int j = i + step - 1; dp[i][j] = s.substr(i, step); for (int k = i; k < j; ++k) { string left = dp[i][k], right = dp[k + 1][j]; if (left.size() + right.size() < dp[i][j].size()) { dp[i][j] = left + right; } } string t = s.substr(i, j - i + 1), replace = ""; auto pos = (t + t).find(t, 1); if (pos >= t.size()) replace = t; else replace = to_string(t.size() / pos) + '[' + dp[i][i + pos - 1] + ']'; if (replace.size() < dp[i][j].size()) dp[i][j] = replace; } } return dp[0][n - 1]; } };
根据热心网友iffalse的留言，我们可以优化上面的方法。如果t是重复的，是不是就不需要再看left.size() + right.size() < dp[i][j].size()了。例如t是abcabcabcabcabc, 最终肯定是5[abc]，不需要再看3[abc]+abcabc或者abcabc+3[abc]。对于一个本身就重复的字符串，最小的长度肯定是n[REPEATED]，不会是某个left+right。所以应该把k的那个循环放在t和replace那部分代码的后面。这样的确提高了一些运算效率的，参见代码如下：
解法二：
class Solution { public: string encode(string s) { int n = s.size(); vector<vector<string>> dp(n, vector<string>(n, "")); for (int step = 1; step <= n; ++step) { for (int i = 0; i + step - 1 < n; ++i) { int j = i + step - 1; dp[i][j] = s.substr(i, step); string t = s.substr(i, j - i + 1), replace = ""; auto pos = (t + t).find(t, 1); if (pos < t.size()) { replace = to_string(t.size() / pos) + "[" + dp[i][i + pos - 1] + "]"; if (replace.size() < dp[i][j].size()) dp[i][j] = replace; continue; } for (int k = i; k < j; ++k) { string left = dp[i][k], right = dp[k + 1][j]; if (left.size() + right.size() < dp[i][j].size()) { dp[i][j] = left + right; } } } } return dp[0][n - 1]; } };
类似题目：
参考资料：
https://leetcode.com/problems/encode-string-with-shortest-length/
