[LintCode] Segment Tree Build II 建立线段树之二-阿里云开发者社区

[LintCode] Segment Tree Build II 建立线段树之二

2018-01-09 1503

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简介：

The structure of Segment Tree is a binary tree which each node has two attributes startand 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 bybuild method.
The left child of node A hasstart=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.

Have you met this question in a real interview?

Yes

Clarification

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

Example

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)

这道题是之前那道Segment Tree Build的拓展，这里面给线段树又增添了一个max变量，然后让我们用一个数组取初始化线段树，其中每个节点的max为该节点start和end代表的数组的坐标区域中的最大值。建树的方法跟之前那道没有什么区别，都是用递归来建立，不同的地方就是在于处理max的时候，如果start小于end，说明该节点还可以继续往下分为左右子节点，那么当前节点的max就是其左右子节点的max的较大值，如果start等于end，说明该节点已经不能继续分了，那么max赋为A[left]即可，参见代码如下：

class Solution {
public:
    /**
     *@param A: a list of integer
     *@return: The root of Segment Tree
     */
    SegmentTreeNode * build(vector<int>& A) {
        return build(A, 0, A.size() - 1);
    }
    SegmentTreeNode* build(vector<int>& A, int start, int end) {
        if (start > end) return NULL;
        SegmentTreeNode *node = new SegmentTreeNode(start, end);
        if (start < end) {
            node->left = build(A, start, (start + end) / 2);
            node->right = build(A, (start + end) / 2 + 1, end);
            node->max = max(node->left->max, node->right->max);
        } else {
            node->max = A[start];
        }
        return node;
    }
};