[LintCode] Segment Tree Modify

CoffeX 發(fā)布于2019-08-14 16:23 / 2643人閱讀

摘要：和其它題目一樣，依然是遞歸的做法。注意若樹的結(jié)點(diǎn)范圍為，那么至少有個(gè)數(shù)在左子樹上，所以語句里用了號。另外，最后一句是調(diào)用遞歸之后的結(jié)果，必須寫在最后面。

Problem

For a Maximum Segment Tree, which each node has an extra value max to store the maximum value in this node"s interval.

Implement a modify function with three parameter root, index and value to change the node"s value with [start, end] = [index, index] to the new given value. Make sure after this change, every node in segment tree still has the max attribute with the correct value.

Example

For segment tree:

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

if call modify(root, 2, 4), we can get:

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

or call modify(root, 4, 0), we can get:

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

Challenge

Do it in O(h) time, h is the height of the segment tree.

Note

和segment tree其它題目一樣，依然是遞歸的做法。注意：若樹的結(jié)點(diǎn)范圍為0~n，那么至少有n/2個(gè)數(shù)在左子樹上，所以if語句里用了<=號。另外，最后一句root.max = Math.max(root.left.max, root.right.max)是調(diào)用遞歸之后的結(jié)果，必須寫在最后面。

Solution

public class Solution {
    public void modify(SegmentTreeNode root, int index, int value) {
        if (index > root.end || index < root.start) return;
        if (root.start == root.end) {
            root.max = value;
            return;
        }
        if (index <= (root.start + root.end) / 2) modify(root.left, index, value);
        else modify(root.right, index, value);
        root.max = Math.max(root.left.max, root.right.max);
    }
}