lintcode ：Segmemt Tree Build II

题目

Segmemt Tree Build II

The structure of Segment Tree is a binary tree which each node has two attributes start and end denote an segment / interval.

start and end are both integers, they should be assigned in following rules:

The root's start and end is given by build method.
The left child of node A has start=A.left, end=(A.left + A.right) / 2.
The right child of node A hasstart=(A.left + A.right) / 2 + 1, end=A.right.
if start equals to end, there will be no children for this node.

Implement a build method with a given array, so that we can create a corresponding segment tree with every node value represent the corresponding interval max value in the array, return the root of this segment tree.

样例

Given [3,2,1,4]. The segment tree will be:

                 [0,  3] (max = 4)
                  /            
        [0,  1] (max = 3)     [2, 3]  (max = 4)
        /                       /             
[0, 0](max = 3)  [1, 1](max = 2)[2, 2](max = 1) [3, 3] (max = 4)

说明

Segment Tree (a.k.a Interval Tree) is an advanced data structure which can support queries like:

which of these intervals contain a given point
which of these points are in a given interval

See wiki: Segment Tree Interval Tree

解题

理解题意：根据给的数组构建段树，该节点有区间及其该区间的最大值组成。区间的左右节点利用上面的规则计算。

/**
 * Definition of SegmentTreeNode:
 * public class SegmentTreeNode {
 *     public int start, end, max;
 *     public SegmentTreeNode left, right;
 *     public SegmentTreeNode(int start, int end, int max) {
 *         this.start = start;
 *         this.end = end;
 *         this.max = max
 *         this.left = this.right = null;
 *     }
 * }
 */

这个节点定义要好好理解。

/**
 * Definition of SegmentTreeNode:
 * public class SegmentTreeNode {
 *     public int start, end, max;
 *     public SegmentTreeNode left, right;
 *     public SegmentTreeNode(int start, int end, int max) {
 *         this.start = start;
 *         this.end = end;
 *         this.max = max
 *         this.left = this.right = null;
 *     }
 * }
 */
public class Solution {
    /**
     *@param A: a list of integer
     *@return: The root of Segment Tree
     */
    public SegmentTreeNode build(int[] A) {
        // write your code here
        return build(0,A.length-1,A);
    }
    public SegmentTreeNode build(int start,int end,int[] A){
        if(start > end ){
            return null;
        }
        SegmentTreeNode root = new SegmentTreeNode(start,end);
        if( start != end){
            int mid = (start + end)/2;
            root.left = build(start,mid,A);
            root.right = build(mid+1,end,A);
            root.max = Math.max(root.left.max,root.right.max);
        }else{
            root.max = A[start];
        }
        return root;
    }
}

Java Code

总耗时: 2532 ms

"""
Definition of SegmentTreeNode:
class SegmentTreeNode:
    def __init__(self, start, end, max):
        self.start, self.end, self.max = start, end, max
        self.left, self.right = None, None
"""

class Solution:    
    # @oaram A: a list of integer
    # @return: The root of Segment Tree
    def build(self, A):
        # write your code here
        return self.buildX(0,len(A) - 1,A)
    def buildX(self,start,end,A):
        if start > end:
            return None
        maxX = 0
        root = SegmentTreeNode(start,end)
        if start != end:
            mid = int((start + end)/2)
            root.left = self.buildX(start,mid,A)
            root.right = self.buildX(mid+1,end,A)
            root.max = max(root.left.max,root.right.max)
        else:
            root.max = A[start]
        return root

Python Code

总耗时: 750 ms