850. Rectangle Area II #
题目 #
We are given a list of (axis-aligned) rectangles
. Each rectangle[i] = [x1, y1, x2, y2]
, where (x1, y1) are the coordinates of the bottom-left corner, and (x2, y2) are the coordinates of the top-right corner of the i
th rectangle.
Find the total area covered by all rectangles
in the plane. Since the answer may be too large, return it modulo 10^9 + 7.
Example 1:
Input: [[0,0,2,2],[1,0,2,3],[1,0,3,1]]
Output: 6
Explanation: As illustrated in the picture.
Example 2:
Input: [[0,0,1000000000,1000000000]]
Output: 49
Explanation: The answer is 10^18 modulo (10^9 + 7), which is (10^9)^2 = (-7)^2 = 49.
Note:
1 <= rectangles.length <= 200
rectanges[i].length = 4
0 <= rectangles[i][j] <= 10^9
- The total area covered by all rectangles will never exceed
2^63 - 1
and thus will fit in a 64-bit signed integer.
题目大意 #
我们给出了一个(轴对齐的)矩形列表 rectangles。 对于 rectangle[i] = [x1, y1, x2, y2],其中(x1,y1)是矩形 i 左下角的坐标,(x2,y2)是该矩形右上角的坐标。找出平面中所有矩形叠加覆盖后的总面积。由于答案可能太大,请返回它对 10 ^ 9 + 7 取模的结果。
提示:
- 1 <= rectangles.length <= 200
- rectanges[i].length = 4
- 0 <= rectangles[i][j] <= 10^9
- 矩形叠加覆盖后的总面积不会超越 2^63 - 1 ,这意味着可以用一个 64 位有符号整数来保存面积结果。
解题思路 #
在二维坐标系中给出一些矩形,要求这些矩形合并之后的面积。由于矩形有重叠,所以需要考虑合并以后的面积。矩形的坐标值也会很大。
这一题给人的感觉很像第 218 题,求天际线的过程也是有楼挡楼,重叠的情况。不过那一题只用求天际线的拐点,所以我们可以对区间做“右边界减一”的处理,防止两个相邻区间因为共点,而导致结果错误。但是这一题如果还是用相同的做法,就会出错,因为“右边界减一”以后,面积会少一部分,最终得到的结果也是偏小的。所以这一题要将线段树改造一下。
思路是先讲 Y 轴上的坐标离线化,转换成线段树。将矩形的 2 条边变成扫描线,左边是入边,右边是出边。
再从左往右遍历每条扫描线,并对 Y 轴上的线段树进行 update。X 轴上的每个坐标区间 * query 线段树总高度的结果 = 区间面积。最后将 X 轴对应的每个区间面积加起来,就是最终矩形合并以后的面积。如下图中间的图。
需要注意的一点是,每次 query 的结果并不一定是连续线段。如上图最右边的图,中间有一段是可能出现镂空的。这种情况看似复杂,其实很简单,因为每段线段树的线段代表的权值高度是不同的,每次 query 最大高度得到的结果已经考虑了中间可能有镂空的情况了。
具体做法,先把各个矩形在 Y 轴方向上离散化,这里的线段树叶子节点不再是一个点了,而是一个区间长度为 1 的区间段。
每个叶子节点也不再是存储一个 int 值,而是存 2 个值,一个是 count 值,用来记录这条区间被覆盖的次数,另一个值是 val 值,用来反映射该线段长度是多少,因为 Y 轴被离散化了,区间坐标间隔都是 1,但是实际 Y 轴的高度并不是 1 ,所以用 val 来反映射原来的高度。
初始化线段树,叶子节点的 count = 0,val 根据题目给的 Y 坐标进行计算。
从左往右遍历每个扫描线。每条扫面线都把对应 update 更新到叶子节点。pushUp 的时候需要合并每个区间段的高度 val 值。如果有区间没有被覆盖,那么这个区间高度 val 为 0,这也就处理了可能“中间镂空”的情况。
func (sat *SegmentAreaTree) pushUp(treeIndex, leftTreeIndex, rightTreeIndex int) { newCount, newValue := sat.merge(sat.tree[leftTreeIndex].count, sat.tree[rightTreeIndex].count), 0 if sat.tree[leftTreeIndex].count > 0 && sat.tree[rightTreeIndex].count > 0 { newValue = sat.merge(sat.tree[leftTreeIndex].val, sat.tree[rightTreeIndex].val) } else if sat.tree[leftTreeIndex].count > 0 && sat.tree[rightTreeIndex].count == 0 { newValue = sat.tree[leftTreeIndex].val } else if sat.tree[leftTreeIndex].count == 0 && sat.tree[rightTreeIndex].count > 0 { newValue = sat.tree[rightTreeIndex].val } sat.tree[treeIndex] = SegmentItem{count: newCount, val: newValue} }
扫描每一个扫描线,先 pushDown 到叶子节点,再 pushUp 到根节点。
遍历到倒数第 2 根扫描线的时候就能得到结果了。因为最后一根扫描线 update 以后,整个线段树全部都归为初始化状态了。
这一题是线段树扫面线解法的经典题。
代码 #
package leetcode
import (
"sort"
)
func rectangleArea(rectangles [][]int) int {
sat, res := SegmentAreaTree{}, 0
posXMap, posX, posYMap, posY, lines := discretization850(rectangles)
tmp := make([]int, len(posYMap))
for i := 0; i < len(tmp)-1; i++ {
tmp[i] = posY[i+1] - posY[i]
}
sat.Init(tmp, func(i, j int) int {
return i + j
})
for i := 0; i < len(posY)-1; i++ {
tmp[i] = posY[i+1] - posY[i]
}
for i := 0; i < len(posX)-1; i++ {
for _, v := range lines[posXMap[posX[i]]] {
sat.Update(posYMap[v.start], posYMap[v.end], v.state)
}
res += ((posX[i+1] - posX[i]) * sat.Query(0, len(posY)-1)) % 1000000007
}
return res % 1000000007
}
func discretization850(positions [][]int) (map[int]int, []int, map[int]int, []int, map[int][]LineItem) {
tmpXMap, tmpYMap, posXArray, posXMap, posYArray, posYMap, lines := map[int]int{}, map[int]int{}, []int{}, map[int]int{}, []int{}, map[int]int{}, map[int][]LineItem{}
for _, pos := range positions {
tmpXMap[pos[0]]++
tmpXMap[pos[2]]++
}
for k := range tmpXMap {
posXArray = append(posXArray, k)
}
sort.Ints(posXArray)
for i, pos := range posXArray {
posXMap[pos] = i
}
for _, pos := range positions {
tmpYMap[pos[1]]++
tmpYMap[pos[3]]++
tmp1 := lines[posXMap[pos[0]]]
tmp1 = append(tmp1, LineItem{start: pos[1], end: pos[3], state: 1})
lines[posXMap[pos[0]]] = tmp1
tmp2 := lines[posXMap[pos[2]]]
tmp2 = append(tmp2, LineItem{start: pos[1], end: pos[3], state: -1})
lines[posXMap[pos[2]]] = tmp2
}
for k := range tmpYMap {
posYArray = append(posYArray, k)
}
sort.Ints(posYArray)
for i, pos := range posYArray {
posYMap[pos] = i
}
return posXMap, posXArray, posYMap, posYArray, lines
}
// LineItem define
type LineItem struct { // 垂直于 x 轴的线段
start, end, state int // state = 1 代表进入,-1 代表离开
}
// SegmentItem define
type SegmentItem struct {
count int
val int
}
// SegmentAreaTree define
type SegmentAreaTree struct {
data []int
tree []SegmentItem
left, right int
merge func(i, j int) int
}
// Init define
func (sat *SegmentAreaTree) Init(nums []int, oper func(i, j int) int) {
sat.merge = oper
data, tree := make([]int, len(nums)), make([]SegmentItem, 4*len(nums))
for i := 0; i < len(nums); i++ {
data[i] = nums[i]
}
sat.data, sat.tree = data, tree
if len(nums) > 0 {
sat.buildSegmentTree(0, 0, len(nums)-1)
}
}
// 在 treeIndex 的位置创建 [left....right] 区间的线段树
func (sat *SegmentAreaTree) buildSegmentTree(treeIndex, left, right int) {
if left == right-1 {
sat.tree[treeIndex] = SegmentItem{count: 0, val: sat.data[left]}
return
}
midTreeIndex, leftTreeIndex, rightTreeIndex := left+(right-left)>>1, sat.leftChild(treeIndex), sat.rightChild(treeIndex)
sat.buildSegmentTree(leftTreeIndex, left, midTreeIndex)
sat.buildSegmentTree(rightTreeIndex, midTreeIndex, right)
sat.pushUp(treeIndex, leftTreeIndex, rightTreeIndex)
}
func (sat *SegmentAreaTree) pushUp(treeIndex, leftTreeIndex, rightTreeIndex int) {
newCount, newValue := sat.merge(sat.tree[leftTreeIndex].count, sat.tree[rightTreeIndex].count), 0
if sat.tree[leftTreeIndex].count > 0 && sat.tree[rightTreeIndex].count > 0 {
newValue = sat.merge(sat.tree[leftTreeIndex].val, sat.tree[rightTreeIndex].val)
} else if sat.tree[leftTreeIndex].count > 0 && sat.tree[rightTreeIndex].count == 0 {
newValue = sat.tree[leftTreeIndex].val
} else if sat.tree[leftTreeIndex].count == 0 && sat.tree[rightTreeIndex].count > 0 {
newValue = sat.tree[rightTreeIndex].val
}
sat.tree[treeIndex] = SegmentItem{count: newCount, val: newValue}
}
func (sat *SegmentAreaTree) leftChild(index int) int {
return 2*index + 1
}
func (sat *SegmentAreaTree) rightChild(index int) int {
return 2*index + 2
}
// 查询 [left....right] 区间内的值
// Query define
func (sat *SegmentAreaTree) Query(left, right int) int {
if len(sat.data) > 0 {
return sat.queryInTree(0, 0, len(sat.data)-1, left, right)
}
return 0
}
func (sat *SegmentAreaTree) queryInTree(treeIndex, left, right, queryLeft, queryRight int) int {
midTreeIndex, leftTreeIndex, rightTreeIndex := left+(right-left)>>1, sat.leftChild(treeIndex), sat.rightChild(treeIndex)
if left > queryRight || right < queryLeft { // segment completely outside range
return 0 // represents a null node
}
if queryLeft <= left && queryRight >= right { // segment completely inside range
if sat.tree[treeIndex].count > 0 {
return sat.tree[treeIndex].val
}
return 0
}
if queryLeft > midTreeIndex {
return sat.queryInTree(rightTreeIndex, midTreeIndex, right, queryLeft, queryRight)
} else if queryRight <= midTreeIndex {
return sat.queryInTree(leftTreeIndex, left, midTreeIndex, queryLeft, queryRight)
}
// merge query results
return sat.merge(sat.queryInTree(leftTreeIndex, left, midTreeIndex, queryLeft, midTreeIndex),
sat.queryInTree(rightTreeIndex, midTreeIndex, right, midTreeIndex, queryRight))
}
// Update define
func (sat *SegmentAreaTree) Update(updateLeft, updateRight, val int) {
if len(sat.data) > 0 {
sat.updateInTree(0, 0, len(sat.data)-1, updateLeft, updateRight, val)
}
}
func (sat *SegmentAreaTree) updateInTree(treeIndex, left, right, updateLeft, updateRight, val int) {
midTreeIndex, leftTreeIndex, rightTreeIndex := left+(right-left)>>1, sat.leftChild(treeIndex), sat.rightChild(treeIndex)
if left > right || left >= updateRight || right <= updateLeft { // 由于叶子节点的区间不在是 left == right 所以这里判断需要增加等号的判断
return // out of range. escape.
}
if updateLeft <= left && right <= updateRight { // segment is fully within update range
if left == right-1 {
sat.tree[treeIndex].count = sat.merge(sat.tree[treeIndex].count, val)
}
if left != right-1 { // update lazy[] for children
sat.updateInTree(leftTreeIndex, left, midTreeIndex, updateLeft, updateRight, val)
sat.updateInTree(rightTreeIndex, midTreeIndex, right, updateLeft, updateRight, val)
sat.pushUp(treeIndex, leftTreeIndex, rightTreeIndex)
}
return
}
sat.updateInTree(leftTreeIndex, left, midTreeIndex, updateLeft, updateRight, val)
sat.updateInTree(rightTreeIndex, midTreeIndex, right, updateLeft, updateRight, val)
// merge updates
sat.pushUp(treeIndex, leftTreeIndex, rightTreeIndex)
}