Google – Search in Interval


Google – Search in Interval
input一堆intervals, 有叠加,给一个值,查询在不在这堆intervals里。(查询会调用很多次)
Follow up是查询有多少个intervals包含这个值,也一样会调用很多次。
[Solution]
__[Solution #1]__
Interval Tree
constructor: buildtree O(nlogn)
queryExist O(logn)
queryNumber O(n)
__[Solution #2]__
Scan-Line
constructor: sort O(nlogn)
queryExist O(logn)
queryNumber O(logn)
思路就是把interval变成两条Line,根据坐标排序,从左到右扫一遍,碰到start就cnt++,碰到end就cnt–。每条Line存当前的cnt。
那么query的时候就用binary search找idx的ceiling,找到之后要分情况考虑:
  1. if ceilling = 0,return 0
  2. else if ceiling – 1 是end, 并且ceiling – 1正好等于idx,那要返回ceiling-1的cnt+1
  3. else 直接返回ceiling – 1的cnt
注意这里的ceiling是第一个大于idx的值,而不是大于等于。

class Interval {
  int start;
  int end;
 
  Interval(int start, int end) {
    this.start = start;
    this.end = end;
  }
}
 
// interval tree solution
class Solution {
 
  TLine root;
 
  public Solution(Interval[] intervals) {
    this.root = new TLine(intervals[0].start, intervals[0].end);
    for (int i = 1; i < intervals.length; i++) {
      insert(root, intervals[i]);
    }
  }
 
  private TLine insert(TLine curr, Interval interval) {
    if (curr == null) {
      return new TLine(interval.start, interval.end);
    }
 
    if (interval.start < curr.l) {
      curr.left = insert(curr.left, interval);
    }
    else {
      curr.right = insert(curr.right, interval);
    }
 
    if (curr.left != null) {
      curr.maxValue = Math.max(curr.maxValue, curr.left.maxValue);
    }
    if (curr.right != null) {
      curr.maxValue = Math.max(curr.maxValue, curr.right.maxValue);
    }
 
    return curr;
  }
 
  // still O(n) time
  public int query(int idx) {
    return dfs(root, idx);
  }
 
  private int dfs(TLine curr, int idx) {
    if (curr == null) {
      return 0;
    }
    int result = 0;
    if (curr.l <= idx && curr.r >= idx) {
      result = 1;
    }
 
    if (curr.left != null && idx <= curr.left.maxValue) {
      return result + dfs(curr.left, idx) + dfs(curr.right, idx);
    }
    else {
      return result + dfs(curr.right, idx);
    }
 
  }
 
  private class TLine {
    int l;
    int r;
    int maxValue;
 
    TLine left;
    TLine right;
 
    TLine(int l, int r) {
      this.l = l;
      this.r = r;
      this.maxValue = r;
    }
  }
}
 
 
// scan-line algorithm
class Solution2 {
 
  List<Line> list = new ArrayList<>();
 
  public Solution2(Interval[] intervals) {
    for (Interval interval : intervals) {
      list.add(new Line(interval.start, true));
      list.add(new Line(interval.end, false));
    }
 
    Collections.sort(list, new Comparator<Line>() {
      public int compare(Line a, Line b) {
        if (a.x == b.x) {
          return a.isStart? -1 : 1;
        }
 
        return a.x - b.x;
      }
    });
 
    int cnt = 0;
    for (int i = 0; i < list.size(); i++) {
      if (list.get(i).isStart) {
        cnt++;
      }
      else {
        cnt--;
      }
      list.get(i).cnt = cnt;
    }
  }
 
  public int query(int idx) {
    int ceil = findCeil(idx, 0, list.size() - 1);
    if (ceil == 0) {
      return 0;
    }
    if (!list.get(ceil - 1).isStart && list.get(ceil - 1).x == idx) {
      return list.get(ceil - 1).cnt + 1;
    }
    return list.get(ceil - 1).cnt;
  }
 
  private int findCeil(int idx, int l, int r) {
    if (idx < list.get(l).x) {
      return l;
    }
    if (list.get(r).x <= idx) {
      return r + 1;
    }
 
    int mid = l + (r - l) / 2;
    if (list.get(mid).x <= idx) {
      if (mid != r && list.get(mid + 1).x > idx) {
        return mid + 1;
      }
      else {
        return findCeil(idx, mid + 1, r);
      }
    }
    else {
      if (mid != l && list.get(mid - 1).x <= idx) {
        return mid;
      }
      else {
        return findCeil(idx, l, mid - 1);
      }
    }
  }
 
  private class Line {
    int x;
    boolean isStart;
    int cnt = 0;
 
    Line(int x, boolean isStart) {
      this.x = x;
      this.isStart = isStart;
    }
  }
}
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