Decode String LeetCode solution in Java

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class Solution {
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  public String decodeString(String s) {
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    int bb = s.indexOf('['); // location of beginning bracket
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    int nbb = s.indexOf('[', bb + 1); // location of next beginning bracket
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    int eb = s.indexOf(']'); // location of ending bracket
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    int n = 0; // number of times to repeat
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    int nl = 1; // number of digits for n
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    char nd; // next digit
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    String insert = ""; // repeated string
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    String end = ""; // remainder of string after repeated portion
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    while (bb != -1) { // while the string contains a beginning bracket
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      while (nbb < eb
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          && nbb > bb) { // while the next beginning bracket is before the ending bracket
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        bb = nbb; // update location of beginning bracket
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        nbb = s.indexOf('[', bb + 1); // update location of next beginning bracket
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      }
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      nl = 1; // reset length of n
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      while (bb - nl >= 0) { // while there are characters in front of the beginning bracket
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        nd = s.charAt(bb - nl); // next digit
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        if (nd <= '9' && nd >= '0') { // if next digit is an integer
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          n += (int) (nd - '0') * Math.pow(10, nl - 1); // update value of n
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          nl++; // increment length of n
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        } else break; // not an integer
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      }
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      insert = s.substring(bb + 1, eb); // set repeated string
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      end = s.substring(eb + 1); // set remainder of string
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      s = s.substring(0, bb - nl + 1); // remove everything after the digits
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      while (n > 0) {
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        s += insert; // add repeated string n times
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        n--;
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      }
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      s += end; // add remainder of string
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      bb = s.indexOf('['); // new location of beginning bracket
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      nbb = s.indexOf('[', bb + 1); // new location of next beginning bracket
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      eb = s.indexOf(']'); // new location of ending bracket
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    }
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    return s;
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  }
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}