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471.java
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471.java
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/* DP, the formula is
dp[i][j] = min(dp[i][j], dp[i][k] + dp[k+1][j])
Note that you can find the string itself in each loop to be abbreviated.
Complexity: O(n^4) replaceAll requires O(n) */
public class Solution {
public String encode(String s) {
String[][] dp = new String[s.length()][s.length()];
for (int l = 0; l < s.length(); l++) {
for (int i = 0; i < s.length() - l; i++) {
int j = i + l;
String substr = s.substring(i, j + 1);
if (l < 4) dp[i][j] = substr;
else {
dp[i][j] = substr;
for (int k = i; k < j; k++) {
if ((dp[i][k] + dp[k + 1][j]).length() < dp[i][j].length()) dp[i][j] = dp[i][k] + dp[k + 1][j];
}
for (int k = 0; k < l; k++) {
String repeatStr = substr.substring(0, k + 1);
if (repeatStr != null && substr.length() % repeatStr.length() == 0
&& substr.replaceAll(repeatStr, "").length() == 0) {
String ss = substr.length() / repeatStr.length() + "[" + dp[i][i+k] + "]";
if(ss.length() < dp[i][j].length()) {
dp[i][j] = ss;
}
}
}
}
}
}
return dp[0][s.length()-1];
}
}