Maximum Number of Visible Points LeetCode solution in Java

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class Solution {
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  public int visiblePoints(List<List<Integer>> points, int angle, List<Integer> location) {
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    int overlap = 0;
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    List<Double> list = new ArrayList<>(points.size());
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    for (List<Integer> p : points) {
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      if (p.get(0) == location.get(0) && p.get(1) == location.get(1)) {
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        overlap++;
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      } else {
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        list.add(angle(p.get(1) - location.get(1), p.get(0) - location.get(0)));
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      }
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    }
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    Collections.sort(list);
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    int max = 0;
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    int n = list.size();
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    int i2 = 0;
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    // list.get(i1) is first angle leg
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    // list.get(i2) is second angle leg
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    for (int i1 = 0; i1 < n; i1++) {
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      // let's grow i1-i2 angle as much as possible
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      // edge case example: angle = 30, i1 = 350 degrees, i2 = 10 degrees
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      // edge case handling: allow i2 to circle around and calculate second leg as (360 +
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      // list.get(i2 % n))
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      // then i1 = 350, i2 = 370, delta = 20 degrees < 30 degrees
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      while ((i2 < n && list.get(i2) - list.get(i1) <= angle)
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          || (i2 >= n && 360 + list.get(i2 % n) - list.get(i1) <= angle)) {
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        i2++;
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      }
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      // after i2 went as far as possible away from i1 under allowed limit - check if a new maximum
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      // found
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      max = Math.max(max, i2 - i1);
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    }
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    return max + overlap;
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  }
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  private double angle(int dy, int dx) {
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    double a = Math.toDegrees(Math.atan2(dy, dx));
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    return (a < 0 ? a + 360 : a);
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  }
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}