刷题前唠嗑

LeetCode? 启动！！！

嗯？怎么是 hard，好长，可恶，看不懂，怎么办

题目：Range 模块

题目链接：715. Range 模块

题目描述

代码与解题思路

今天是个好日子（毕竟是周日），必须露两手，来看代码：

const N int = 1e9

type node struct {
	lch   *node
	rch   *node
	added bool
	lazy  int
}

type segmentTree struct {
	root *node
}

func newSegmentTree() *segmentTree {
	return &segmentTree{
		root: new(node),
	}
}

func (t *segmentTree) update(n *node, l, r, i, j, x int) {
	if l >= i && r <= j {
		n.added = x == 1
		n.lazy = x
		return
	}
	t.pushdown(n)
	m := int(uint(l+r) >> 1)
	if i <= m {
		t.update(n.lch, l, m, i, j, x)
	}
	if j > m {
		t.update(n.rch, m+1, r, i, j, x)
	}
	t.pushup(n)
}

func (t *segmentTree) query(n *node, l, r, i, j int) bool {
	if l >= i && r <= j {
		return n.added
	}
	t.pushdown(n)
	v := true
	m := int(uint(l+r) >> 1)
	if i <= m {
		v = v && t.query(n.lch, l, m, i, j)
	}
	if j > m {
		v = v && t.query(n.rch, m+1, r, i, j)
	}
	return v
}

func (t *segmentTree) pushup(n *node) {
	n.added = n.lch.added && n.rch.added
}

func (t *segmentTree) pushdown(n *node) {
	if n.lch == nil {
		n.lch = new(node)
	}
	if n.rch == nil {
		n.rch = new(node)
	}
	if n.lazy != 0 {
		n.lch.added = n.lazy == 1
		n.rch.added = n.lazy == 1
		n.lch.lazy = n.lazy
		n.rch.lazy = n.lazy
		n.lazy = 0
	}
}

type RangeModule struct {
	t *segmentTree
}

func Constructor() RangeModule {
	return RangeModule{
		t: newSegmentTree(),
	}
}

func (this *RangeModule) AddRange(left int, right int) {
	this.t.update(this.t.root, 1, N, left, right-1, 1)
}

func (this *RangeModule) QueryRange(left int, right int) bool {
	return this.t.query(this.t.root, 1, N, left, right-1)
}

func (this *RangeModule) RemoveRange(left int, right int) {
	this.t.update(this.t.root, 1, N, left, right-1, -1)
}

/**
 * Your RangeModule object will be instantiated and called as such:
 * obj := Constructor();
 * obj.AddRange(left,right);
 * param_2 := obj.QueryRange(left,right);
 * obj.RemoveRange(left,right);
 */

只需要 5 秒，来看解题流程：

1. 打开一份大佬题解
2. CTRL + C
3. CTRL + V
4. 提交代码
5. 过啦

。。。

今天是线段树，真不会，没办法，只好使用 CV 大法了