zoukankan      html  css  js  c++  java
  • Lintcode: Segment Tree Query II

    For an array, we can build a SegmentTree for it, each node stores an extra attribute count to denote the number of elements in the the array which value is between interval start and end. (The array may not fully filled by elements)
    
    Design a query method with three parameters root, start and end, find the number of elements in the in array's interval [start, end] by the given root of value SegmentTree.
    
    Have you met this question in a real interview? Yes
    Example
    For array [0, 2, 3], the corresponding value Segment Tree is:
    
                         [0, 3, count=3]
                         /             
              [0,1,count=1]             [2,3,count=2]
              /                        /            
       [0,0,count=1] [1,1,count=0] [2,2,count=1], [3,3,count=1]
    query(1, 1), return 0
    
    query(1, 2), return 1
    
    query(2, 3), return 2
    
    query(0, 2), return 2
    
    Note
    It is much easier to understand this problem if you finished Segment Tree Buildand Segment Tree Query first.

    注意23行

     1 /**
     2  * Definition of SegmentTreeNode:
     3  * public class SegmentTreeNode {
     4  *     public int start, end, count;
     5  *     public SegmentTreeNode left, right;
     6  *     public SegmentTreeNode(int start, int end, int count) {
     7  *         this.start = start;
     8  *         this.end = end;
     9  *         this.count = count;
    10  *         this.left = this.right = null;
    11  *     }
    12  * }
    13  */
    14 public class Solution {
    15     /**
    16      *@param root, start, end: The root of segment tree and 
    17      *                         an segment / interval
    18      *@return: The count number in the interval [start, end]
    19      */
    20     public int query(SegmentTreeNode root, int start, int end) {
    21         // write your code here
    22         if (root == null || root.end < start || root.start > end) return 0;
    23         if (root.start>=start && root.end<=end) return root.count;
    24         int mid = (root.start + root.end)/2;
    25         if (end <= mid) return query(root.left, start, end);
    26         else if (start > mid) return query(root.right, start, end);
    27         else return query(root.left, start, mid) + query(root.right, mid+1, end);
    28     }
    29 }
  • 相关阅读:
    关于重构的一些方法
    java基础 逻辑
    java基础
    去重和数组排序
    表单验证
    JS实例5
    window.document对象
    JS实例4
    JS实例3
    JS实例2
  • 原文地址:https://www.cnblogs.com/EdwardLiu/p/5174965.html
Copyright © 2011-2022 走看看