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Commit 5779906

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clean code
1 parent dd5beed commit 5779906

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6 files changed

+98
-48
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6 files changed

+98
-48
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‎leetcode/leetcode-09/src/main/java/Solution891.java

Lines changed: 1 addition & 1 deletion
Original file line numberDiff line numberDiff line change
@@ -26,7 +26,7 @@ public int sumSubseqWidths(int[] nums) {
2626
https://leetcode.cn/problems/sum-of-subsequence-widths/description/
2727
2828
一个序列的 宽度 定义为该序列中最大元素和最小元素的差值。
29-
给你一个整数数组 nums ,返回 nums 的所有非空 子序列 的 宽度之和 。由于答案可能非常大,请返回对 109 + 7 取余 后的结果。
29+
给你一个整数数组 nums ,返回 nums 的所有非空 子序列 的 宽度之和 。由于答案可能非常大,请返回对 10^9 + 7 取余 后的结果。
3030
子序列 定义为从一个数组里删除一些(或者不删除)元素,但不改变剩下元素的顺序得到的数组。例如,[3,6,2,7] 就是数组 [0,3,1,6,2,2,7] 的一个子序列。
3131
提示:
3232
1 <= nums.length <= 10^5

‎leetcode/leetcode-18/src/main/java/Solution1735.java

Lines changed: 1 addition & 1 deletion
Original file line numberDiff line numberDiff line change
@@ -59,7 +59,7 @@ static long quickPow(long a, long b) {
5959
1735. 生成乘积数组的方案数
6060
https://leetcode.cn/problems/count-ways-to-make-array-with-product/description/
6161
62-
给你一个二维整数数组 queries ,其中 queries[i] = [ni, ki] 。第 i 个查询 queries[i] 要求构造长度为 ni 、每个元素都是正整数的数组,且满足所有元素的乘积为 ki ,请你找出有多少种可行的方案。由于答案可能会很大,方案数需要对 109 + 7 取余 。
62+
给你一个二维整数数组 queries ,其中 queries[i] = [ni, ki] 。第 i 个查询 queries[i] 要求构造长度为 ni 、每个元素都是正整数的数组,且满足所有元素的乘积为 ki ,请你找出有多少种可行的方案。由于答案可能会很大,方案数需要对 10^9 + 7 取余 。
6363
请你返回一个整数数组 answer,满足 answer.length == queries.length ,其中 answer[i]是第 i 个查询的结果。
6464
提示:
6565
1 <= queries.length <= 10^4

‎leetcode/leetcode-37/src/main/java/Solution3605.java

Lines changed: 1 addition & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -182,4 +182,5 @@ private int getGCD(int num1, int num2) {
182182
二分答案 + ST 表。
183183
时间复杂度 O(nlogU)
184184
二分答案 + logTrick。
185+
rating 2434 (clist.by)
185186
*/
Lines changed: 86 additions & 42 deletions
Original file line numberDiff line numberDiff line change
@@ -1,59 +1,101 @@
1+
import java.util.Arrays;
2+
import java.util.Comparator;
3+
14
public class Solution3608 {
2-
public int minTime(int n, int[][] edges, int k) {
3-
int maxTime = 0;
4-
for (int[] edge : edges) {
5-
if (edge[2] > maxTime) {
6-
maxTime = edge[2];
5+
static class V1 {
6+
public int minTime(int n, int[][] edges, int k) {
7+
int maxTime = 0;
8+
for (int[] edge : edges) {
9+
if (edge[2] > maxTime) {
10+
maxTime = edge[2];
11+
}
712
}
8-
}
913

10-
int left = 0;
11-
int right = maxTime;
12-
while (left < right) {
13-
int mid = left + (right - left) / 2;
14-
// 边界二分 F, F,..., F, [T, T,..., T]
15-
// ----------------------^
16-
if (check(mid, n, edges, k)) {
17-
right = mid;
18-
} else {
19-
left = mid + 1;
14+
int left = 0;
15+
int right = maxTime;
16+
while (left < right) {
17+
int mid = left + (right - left) / 2;
18+
// 边界二分 F, F,..., F, [T, T,..., T]
19+
// ----------------------^
20+
if (check(mid, n, edges, k)) {
21+
right = mid;
22+
} else {
23+
left = mid + 1;
24+
}
2025
}
26+
return left;
2127
}
22-
return left;
23-
}
2428

25-
private boolean check(int mid, int n, int[][] edges, int k) {
26-
DSU dsu = new DSU(n);
27-
for (int[] edge : edges) {
28-
int time = edge[2];
29-
if (time > mid) {
30-
int u = edge[0], v = edge[1];
31-
dsu.union(u, v);
29+
private boolean check(int mid, int n, int[][] edges, int k) {
30+
DSU dsu = new DSU(n);
31+
for (int[] edge : edges) {
32+
int time = edge[2];
33+
if (time > mid) {
34+
int u = edge[0], v = edge[1];
35+
dsu.union(u, v);
36+
}
3237
}
38+
return dsu.sz >= k;
3339
}
34-
return dsu.sz >= k;
35-
}
3640

37-
static class DSU {
38-
int[] fa;
39-
int sz;
41+
static class DSU {
42+
int[] fa;
43+
int sz;
4044

41-
public DSU(int n) {
42-
fa = new int[n];
43-
for (int i = 0; i < n; i++) fa[i] = i;
44-
sz = n;
45+
public DSU(int n) {
46+
fa = new int[n];
47+
for (int i = 0; i < n; i++) fa[i] = i;
48+
sz = n;
49+
}
50+
51+
int find(int x) { // 查找
52+
return x == fa[x] ? fa[x] : (fa[x] = find(fa[x]));
53+
}
54+
55+
void union(int p, int q) { // 合并
56+
p = find(p);
57+
q = find(q);
58+
if (p == q) return;
59+
fa[q] = p;
60+
sz--;
61+
}
4562
}
63+
}
4664

47-
int find(int x) { // 查找
48-
return x == fa[x] ? fa[x] : (fa[x] = find(fa[x]));
65+
static class V2 {
66+
public int minTime(int n, int[][] edges, int k) {
67+
DSU dsu = new DSU(n);
68+
Arrays.sort(edges, Comparator.comparingInt(o -> -o[2]));
69+
for (int[] e : edges) {
70+
dsu.union(e[0], e[1]);
71+
if (dsu.sz < k) {
72+
return e[2];
73+
}
74+
}
75+
return 0;
4976
}
5077

51-
void union(int p, int q) { // 合并
52-
p = find(p);
53-
q = find(q);
54-
if (p == q) return;
55-
fa[q] = p;
56-
sz--;
78+
static class DSU {
79+
int[] fa;
80+
int sz;
81+
82+
public DSU(int n) {
83+
fa = new int[n];
84+
for (int i = 0; i < n; i++) fa[i] = i;
85+
sz = n;
86+
}
87+
88+
int find(int x) { // 查找
89+
return x == fa[x] ? fa[x] : (fa[x] = find(fa[x]));
90+
}
91+
92+
void union(int p, int q) { // 合并
93+
p = find(p);
94+
q = find(q);
95+
if (p == q) return;
96+
fa[q] = p;
97+
sz--;
98+
}
5799
}
58100
}
59101
}
@@ -79,4 +121,6 @@ void union(int p, int q) { // 合并
79121
不存在重复的边。
80122
81123
二分答案 + 并查集。
124+
相似题目: 3613. 最小化连通分量的最大成本
125+
https://leetcode.cn/problems/minimize-maximum-component-cost/description/
82126
*/

‎leetcode/leetcode-37/src/main/java/Solution3609.java

Lines changed: 1 addition & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -74,4 +74,5 @@ public int minMoves(int sx, int sy, int tx, int ty) {
7474
2.3 x = 2y 把 x 减半得到 y,把 x 减去 y 也得到 y,所以是一样的,可以整合到 2.2
7575
相似题目: 780. 到达终点
7676
https://leetcode.cn/problems/reaching-points/
77+
rating 2451 (clist.by)
7778
*/

‎leetcode/leetcode-37/src/test/java/Solution3608Tests.java

Lines changed: 8 additions & 4 deletions
Original file line numberDiff line numberDiff line change
@@ -2,15 +2,17 @@
22
import org.junit.jupiter.api.Test;
33

44
public class Solution3608Tests {
5-
private final Solution3608 solution3608 = new Solution3608();
5+
private final Solution3608.V1 solution3608_v1 = new Solution3608.V1();
6+
private final Solution3608.V2 solution3608_v2 = new Solution3608.V2();
67

78
@Test
89
public void example1() {
910
int n = 2;
1011
int[][] edges = UtUtils.stringToInts2("[[0,1,3]]");
1112
int k = 2;
1213
int expected = 3;
13-
Assertions.assertEquals(expected, solution3608.minTime(n, edges, k));
14+
Assertions.assertEquals(expected, solution3608_v1.minTime(n, edges, k));
15+
Assertions.assertEquals(expected, solution3608_v2.minTime(n, edges, k));
1416
}
1517

1618
@Test
@@ -19,7 +21,8 @@ public void example2() {
1921
int[][] edges = UtUtils.stringToInts2("[[0,1,2],[1,2,4]]");
2022
int k = 3;
2123
int expected = 4;
22-
Assertions.assertEquals(expected, solution3608.minTime(n, edges, k));
24+
Assertions.assertEquals(expected, solution3608_v1.minTime(n, edges, k));
25+
Assertions.assertEquals(expected, solution3608_v2.minTime(n, edges, k));
2326
}
2427

2528
@Test
@@ -28,6 +31,7 @@ public void example3() {
2831
int[][] edges = UtUtils.stringToInts2("[[0,2,5]]");
2932
int k = 2;
3033
int expected = 0;
31-
Assertions.assertEquals(expected, solution3608.minTime(n, edges, k));
34+
Assertions.assertEquals(expected, solution3608_v1.minTime(n, edges, k));
35+
Assertions.assertEquals(expected, solution3608_v2.minTime(n, edges, k));
3236
}
3337
}

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