Monday, August 8, 2016

Google – Find a Path Among Circles


Google – Find a Path Among Circles
有一个通道,在y坐标0到1之间:
y = 1 —————————–
y = 0 —————————–
然后input一堆Circles,每个Circles有三个值,x, y, r, 分别表示圆心和半径。问题是在这条通道里是否能找出一条路径,可以从x的负无穷到正无穷。
路径不一定要直线,可以随便弯弯曲曲,但是如果有一个或者多个circle完全block了通道,就没有路径。
[Mistake]
一开始以为就是道Merge Interval的题目,合并每个圆的纵向interval, [y – r, y + r],一旦发现超过[0, 1]范围就返回false.
但是后来发现不是这样的:
比如一个圆在x = 0处,纵向interval为[0.5, 1.5],
另一个圆在x = 10000处,反正就是离的非常远,然后纵向interval为[-0.5, 0.9]
如果merge的话,肯定是超过[0, 1]范围,但是依然可以找出一条满足要求的路径。
所以这道题merge interval是有条件的,只有当两个圆相交或者相切,才能merge它们的纵向interval。
[Solution]
其实说白了就是cluster circles。只要有一个cluster的纵向interval超过范围了,就返回false。
Merge Intervals的变形,只有相交或者相切才merge,需要O( n2 ) time,因为每处理一个circle,都要和之前处理过的所有circle判断是否相交。
注意不能通过单纯通过sort x来判断是否相交,sort后第i个圆甚至可能出现和第i – 2个圆相交,但是和i – 1的圆没有半毛钱关系的情况。所以暂时没有想到比quadratic time更好的解法。
另外相交相切的条件:
圆心距离 < r1 + r2 相交
圆心距离 = r1 + r2 相切
圆心距离 > r1 + r2 相离
class Circle {
  double x;
  double y;
  double r;
  Circle(double x, double y, double r) {
    this.x = x;
    this.y = y;
    this.r = r;
  }
}
class Solution {
  public boolean hasPath(Circle[] circles) {
    if (circles == null || circles.length == 0) {
      return true;
    }
    Map<Circle, Interval> map = new HashMap<>();
    map.put(circles[0], new Interval(circles[0].y - circles[0].r, circles[0].y + circles[0].r));
    for (int i = 1; i < circles.length; i++) {
      boolean hasIntersection = false;
      Interval newInterval = new Interval(circles[i].y - circles[i].r, circles[i].y + circles[i].r);
      for (int j = 0; j < i; j++) {
        if (isIntersect(circles[j], circles[i])) {
          merge(map.get(circles[j]), newInterval);
          if (map.get(circles[j]).s <= 0 && map.get(circles[j]).e >= 1) {
            return false;
          }
          map.put(circles[i], map.get(circles[j]));
          hasIntersection = true;
        }
      }
// maybe first get all intersect intervals and merge them at same time
// after for loop, update all intervals to the merged interval
      if (!hasIntersection) {
        map.put(circles[i], newInterval);
      }
    }
    return true;
  }
  private boolean isIntersect(Circle a, Circle b) {
    double x = Math.abs(a.x - b.x);
    double y = Math.abs(a.y - b.y);
    double len = Math.sqrt(x * x + y * y);
    return len <= a.r + b.r;
  }
  // If intersect, there must be an overlap
  private void merge(Interval interval, Interval newInterval) {
    interval.s = Math.min(interval.s, newInterval.s);
    interval.e = Math.max(interval.e, newInterval.e);
  }
  private class Interval {
    double s;
    double e;
    Interval(double s, double e) {
      this.s = s;
      this.e = e;
    }
  }
}
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