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Commit d326d43

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clean code & 850 (1)
1 parent 11e11f8 commit d326d43

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10 files changed

+443
-35
lines changed

10 files changed

+443
-35
lines changed

‎leetcode/leetcode-04/src/main/java/Solution386.java

Lines changed: 41 additions & 14 deletions
Original file line numberDiff line numberDiff line change
@@ -2,22 +2,48 @@
22
import java.util.List;
33

44
public class Solution386 {
5-
public List<Integer> lexicalOrder(int n) {
6-
List<Integer> resList = new ArrayList<>();
7-
int num = 1;
8-
// 保证最多不超过 n 个数
9-
for (int i = 0; i < n; i++) {
10-
resList.add(num);
11-
if (num * 10 <= n) {
12-
num *= 10;
13-
} else {
14-
while (num % 10 == 9 || num + 1 > n) {
15-
num /= 10;
5+
// 迭代
6+
static class V1 {
7+
public List<Integer> lexicalOrder(int n) {
8+
List<Integer> ans = new ArrayList<>();
9+
int num = 1;
10+
// 保证最多不超过 n 个数
11+
for (int i = 0; i < n; i++) {
12+
ans.add(num);
13+
if (num * 10 <= n) {
14+
num *= 10;
15+
} else {
16+
while (num % 10 == 9 || num + 1 > n) {
17+
num /= 10;
18+
}
19+
num++;
1620
}
17-
num++;
21+
}
22+
return ans;
23+
}
24+
}
25+
26+
// 递归
27+
static class V2 {
28+
int n;
29+
List<Integer> ans;
30+
31+
public List<Integer> lexicalOrder(int n) {
32+
this.n = n;
33+
ans = new ArrayList<>();
34+
for (int i = 1; i <= 9; i++) {
35+
dfs(i);
36+
}
37+
return ans;
38+
}
39+
40+
private void dfs(int cur) {
41+
if (cur > n) return;
42+
ans.add(cur);
43+
for (int i = 0; i <= 9; i++) {
44+
dfs(cur * 10 + i);
1845
}
1946
}
20-
return resList;
2147
}
2248
}
2349
/*
@@ -29,5 +55,6 @@ public List<Integer> lexicalOrder(int n) {
2955
提示:
3056
1 <= n <= 5 * 10^4
3157
32-
迭代。
58+
迭代 / 递归。
59+
时间复杂度 O(n)。
3360
*/

‎leetcode/leetcode-04/src/test/java/Solution386Tests.java

Lines changed: 6 additions & 3 deletions
Original file line numberDiff line numberDiff line change
@@ -4,19 +4,22 @@
44
import java.util.List;
55

66
public class Solution386Tests {
7-
private final Solution386 solution386 = new Solution386();
7+
private final Solution386.V1 solution386_v1 = new Solution386.V1();
8+
private final Solution386.V2 solution386_v2 = new Solution386.V2();
89

910
@Test
1011
public void example1() {
1112
int n = 13;
1213
List<Integer> expected = List.of(1, 10, 11, 12, 13, 2, 3, 4, 5, 6, 7, 8, 9);
13-
Assertions.assertEquals(expected, solution386.lexicalOrder(n));
14+
Assertions.assertEquals(expected, solution386_v1.lexicalOrder(n));
15+
Assertions.assertEquals(expected, solution386_v2.lexicalOrder(n));
1416
}
1517

1618
@Test
1719
public void example2() {
1820
int n = 2;
1921
List<Integer> expected = List.of(1, 2);
20-
Assertions.assertEquals(expected, solution386.lexicalOrder(n));
22+
Assertions.assertEquals(expected, solution386_v1.lexicalOrder(n));
23+
Assertions.assertEquals(expected, solution386_v2.lexicalOrder(n));
2124
}
2225
}
Lines changed: 144 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,144 @@
1+
import java.util.ArrayList;
2+
import java.util.Arrays;
3+
import java.util.List;
4+
import java.util.Map;
5+
import java.util.TreeMap;
6+
7+
public class Solution850 {
8+
private static final int MOD = (int) (1e9 + 7);
9+
10+
public int rectangleArea(int[][] rectangles) {
11+
int m = 0;
12+
TreeMap<Integer, Integer> mp = new TreeMap<>();
13+
// for (int[] sq : squares) {
14+
// mp.put(sq[0], 1);
15+
// mp.put(sq[0] + sq[2], 1);
16+
// }
17+
for (int[] re : rectangles) {
18+
mp.put(re[0], 1);
19+
mp.put(re[2], 1);
20+
}
21+
for (Map.Entry<Integer, Integer> p : mp.entrySet()) p.setValue(m++);
22+
int[] A = new int[m];
23+
for (Map.Entry<Integer, Integer> p : mp.entrySet()) A[p.getValue()] = p.getKey();
24+
// 离散化结束
25+
26+
// 把正方形的上下边界取出来
27+
List<int[]> vec = new ArrayList<>();
28+
// for (int[] sq : squares) {
29+
// vec.add(new int[]{sq[1], mp.get(sq[0]) + 1, mp.get(sq[0] + sq[2]), 1});
30+
// vec.add(new int[]{sq[1] + sq[2], mp.get(sq[0]) + 1, mp.get(sq[0] + sq[2]), -1});
31+
// }
32+
for (int[] re : rectangles) {
33+
vec.add(new int[]{re[1], mp.get(re[0]) + 1, mp.get(re[2]), 1});
34+
vec.add(new int[]{re[3], mp.get(re[0]) + 1, mp.get(re[2]), -1});
35+
}
36+
vec.sort(Arrays::compare);
37+
38+
// 求总的面积并
39+
long tot = 0;
40+
LazySegmentTree seg = new LazySegmentTree(m);
41+
seg.build(A, 1, 1, m - 1);
42+
for (int i = 0; i + 1 < vec.size(); i++) {
43+
// 考虑水平线 y = vec[i][0] 和 y = vec[i + 1][0] 之间的情况
44+
seg.modify(1, 1, m - 1, vec.get(i)[1], vec.get(i)[2], vec.get(i)[3]);
45+
// 求横截长度
46+
int len = A[m - 1] - A[0];
47+
// 如果最小覆盖数是 0,那么扣掉相应的长度
48+
if (seg.info[1].mn == 0) len -= seg.info[1].len;
49+
// 面积 = 横截长度 * 高度差
50+
tot += (long) len * (vec.get(i + 1)[0] - vec.get(i)[0]);
51+
}
52+
return (int) (tot % MOD);
53+
}
54+
55+
// 线段树模板,只需要实现 mergeInfo 和 _do,其余都是固定的
56+
static class LazySegmentTree {
57+
static class Info {
58+
// mn:当前节点的最小覆盖数
59+
// len:满足覆盖数 = 最小覆盖数的 A[i] 之和
60+
// lazy:加法的懒标记
61+
int mn, len, lazy;
62+
63+
public Info(int mn, int len, int lazy) {
64+
this.mn = mn;
65+
this.len = len;
66+
this.lazy = lazy;
67+
}
68+
}
69+
70+
Info mergeInfo(Info a, Info b) {
71+
int mn = Math.min(a.mn, b.mn);
72+
return new Info(mn, (a.mn == mn ? a.len : 0) + (b.mn == mn ? b.len : 0), 0);
73+
}
74+
75+
// 对节点的覆盖数整个增加 qv,只影响 mn,不影响 len
76+
void _do(int p, int qv) {
77+
info[p].mn += qv;
78+
info[p].lazy += qv;
79+
}
80+
81+
int n;
82+
Info[] info;
83+
84+
public LazySegmentTree(int n) {
85+
this.n = n;
86+
info = new Info[4 * n];
87+
Arrays.fill(info, new Info(0, 0, 0));
88+
}
89+
90+
void build(int[] A, int p, int l, int r) {
91+
if (l == r) {
92+
info[p] = new Info(0, A[r] - A[r - 1], 0);
93+
return;
94+
}
95+
int m = (l + r) >> 1;
96+
build(A, p << 1, l, m);
97+
build(A, p << 1 | 1, m + 1, r);
98+
maintain(p);
99+
}
100+
101+
void maintain(int p) {
102+
info[p] = mergeInfo(info[p << 1], info[p << 1 | 1]);
103+
}
104+
105+
void spread(int p) {
106+
if (info[p].lazy == 0) return;
107+
_do(p << 1, info[p].lazy);
108+
_do(p << 1 | 1, info[p].lazy);
109+
info[p].lazy = 0;
110+
}
111+
112+
void modify(int p, int l, int r, int ql, int qr, int qv) {
113+
if (ql <= l && r <= qr) {
114+
_do(p, qv);
115+
return;
116+
}
117+
spread(p);
118+
int m = (l + r) >> 1;
119+
if (ql <= m) modify(p << 1, l, m, ql, qr, qv);
120+
if (qr > m) modify(p << 1 | 1, m + 1, r, ql, qr, qv);
121+
maintain(p);
122+
}
123+
}
124+
}
125+
/*
126+
850. 矩形面积 II
127+
https://leetcode.cn/problems/rectangle-area-ii/description/
128+
129+
给你一个轴对齐的二维数组 rectangles 。 对于 rectangle[i] = [x1, y1, x2, y2],其中 (xi1, yi1) 是该矩形 左下角 的坐标, (xi2, yi2) 是该矩形 右上角 的坐标。
130+
计算平面中所有 rectangles 所覆盖的 总面积 。任何被两个或多个矩形覆盖的区域应只计算 一次 。
131+
返回 总面积 。因为答案可能太大,返回 10^9 + 7 的 模 。
132+
提示:
133+
1 <= rectangles.length <= 200
134+
rectanges[i].length = 4
135+
0 <= xi1, yi1, xi2, yi2 <= 10^9
136+
xi1 <= xi2
137+
yi1 <= yi2
138+
所有矩阵面积不为 0。
139+
140+
矩形面积并。
141+
Lazy 线段树 + 扫描线。
142+
相似题目: 3454. 分割正方形 II
143+
https://leetcode.cn/problems/separate-squares-ii/description/
144+
*/
Lines changed: 20 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,20 @@
1+
import org.junit.jupiter.api.Assertions;
2+
import org.junit.jupiter.api.Test;
3+
4+
public class Solution850Tests {
5+
private final Solution850 solution850 = new Solution850();
6+
7+
@Test
8+
public void example1() {
9+
int[][] rectangles = UtUtils.stringToInts2("[[0,0,2,2],[1,0,2,3],[1,0,3,1]]");
10+
int expected = 6;
11+
Assertions.assertEquals(expected, solution850.rectangleArea(rectangles));
12+
}
13+
14+
@Test
15+
public void example2() {
16+
int[][] rectangles = UtUtils.stringToInts2("[[0,0,1000000000,1000000000]]");
17+
int expected = 49;
18+
Assertions.assertEquals(expected, solution850.rectangleArea(rectangles));
19+
}
20+
}

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