Ambiguous Coordinates LeetCode solution in Java

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class Solution {
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  public List<String> ret;
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  public List<String> ans;
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  public List<String> ambiguousCoordinates(String s) {
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    ret = new ArrayList<>();
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    ans = new ArrayList<>();
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    String start = s.substring(0, 2);
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    util(s, 1);
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    fun();
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    return ans;
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  }
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  // putting comma
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  void util(String s, int idx) {
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    if (idx == s.length() - 2) {
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      return;
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    }
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    String ns = s.substring(0, idx + 1) + ", " + s.substring(idx + 1);
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    ret.add(ns);
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    util(s, idx + 1);
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  }
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  // helper function for puting decimals after comma
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  void fun() {
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    for (String s : ret) {
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      int cIndex = 0;
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      for (int i = 0; i < s.length(); i++) {
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        if (s.charAt(i) == ',') {
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          cIndex = i;
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          break;
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        }
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      }
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      String a = s.substring(1, cIndex);
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      String b = s.substring(cIndex + 2, s.length() - 1);
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      List<String> n1 = dot(a);
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      List<String> n2 = dot(b);
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      if (n1 == null || n2 == null) { // invalid strings
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        continue;
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      } else { // valid strings
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        for (String fir : n1) {
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          for (String sec : n2) {
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            ans.add("(" + fir + ", " + sec + ")");
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          }
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        }
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      }
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    }
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  }
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  // putting decimal point
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  List<String> dot(String n) {
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    List<String> li = new ArrayList<>();
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    if (n.length() == 1) {
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      li.add(n);
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    } else {
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      // just checking for first and last zeroes and making conditions accordingly
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      if (n.charAt(n.length() - 1) == '0') {
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        if (n.charAt(0) == '0') {
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          return null;
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        } else {
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          li.add(n);
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        }
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      } else if (n.charAt(0) == '0') {
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        li.add("0." + n.substring(1));
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      } else {
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        for (int i = 0; i < n.length() - 1; i++) {
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          li.add(n.substring(0, i + 1) + "." + n.substring(i + 1));
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        }
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        li.add(n); // without any decimal
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      }
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    }
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    return li;
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  }
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}