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Commit fe8a245

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clean code
1 parent bc075d2 commit fe8a245

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32 files changed

+565
-286
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32 files changed

+565
-286
lines changed

‎leetcode/leetcode-01/src/main/java/Solution63.java‎

Lines changed: 5 additions & 14 deletions
Original file line numberDiff line numberDiff line change
@@ -3,27 +3,18 @@ public int uniquePathsWithObstacles(int[][] obstacleGrid) {
33
int m = obstacleGrid.length;
44
int n = obstacleGrid[0].length;
55

6-
if (obstacleGrid[0][0] == 1) {
7-
return 0;
8-
}
9-
6+
if (obstacleGrid[0][0] == 1) return 0;
107
// f[i][j] 表示到达坐标 (i,j) 的路径总数
118
int[][] f = new int[m][n];
129
// 初始状态
1310
f[0][0] = 1;
1411
for (int i = 1; i < m; i++) {
15-
if (obstacleGrid[i][0] == 0) {
16-
f[i][0] = 1;
17-
} else {
18-
break;
19-
}
12+
if (obstacleGrid[i][0] != 0) break;
13+
f[i][0] = 1;
2014
}
2115
for (int j = 1; j < n; j++) {
22-
if (obstacleGrid[0][j] == 0) {
23-
f[0][j] = 1;
24-
} else {
25-
break;
26-
}
16+
if (obstacleGrid[0][j] != 0) break;
17+
f[0][j] = 1;
2718
}
2819
// 状态转移
2920
for (int i = 1; i < m; i++) {

‎leetcode/leetcode-01/src/main/java/Solution86.java‎

Lines changed: 20 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -27,6 +27,26 @@ public ListNode partition(ListNode head, int x) {
2727
}
2828
return dummy.next;
2929
}
30+
31+
public ListNode partition2(ListNode head, int x) {
32+
ListNode small = new ListNode(0);
33+
ListNode smallHead = small;
34+
ListNode large = new ListNode(0);
35+
ListNode largeHead = large;
36+
while (head != null) {
37+
if (head.val < x) {
38+
small.next = head;
39+
small = small.next;
40+
} else {
41+
large.next = head;
42+
large = large.next;
43+
}
44+
head = head.next;
45+
}
46+
large.next = null;
47+
small.next = largeHead.next;
48+
return smallHead.next;
49+
}
3050
}
3151
/*
3252
86. 分隔链表

‎leetcode/leetcode-01/src/test/java/Solution86Tests.java‎

Lines changed: 2 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -10,6 +10,7 @@ public void example1() {
1010
int x = 3;
1111
ListNode expected = ListNode.buildListNode(new int[]{1, 2, 2, 4, 3, 5});
1212
Assertions.assertTrue(ListNode.assertListNodeEquals(expected, solution86.partition(head, x)));
13+
Assertions.assertTrue(ListNode.assertListNodeEquals(expected, solution86.partition2(head, x)));
1314
}
1415

1516
@Test
@@ -18,5 +19,6 @@ public void example2() {
1819
int x = 2;
1920
ListNode expected = ListNode.buildListNode(new int[]{1, 2});
2021
Assertions.assertTrue(ListNode.assertListNodeEquals(expected, solution86.partition(head, x)));
22+
Assertions.assertTrue(ListNode.assertListNodeEquals(expected, solution86.partition2(head, x)));
2123
}
2224
}

‎leetcode/leetcode-02/src/main/java/Solution122.java‎

Lines changed: 37 additions & 1 deletion
Original file line numberDiff line numberDiff line change
@@ -1,5 +1,35 @@
1+
import java.util.Arrays;
2+
13
public class Solution122 {
4+
private int[] prices;
5+
private int[][] memo;
6+
27
public int maxProfit(int[] prices) {
8+
this.prices = prices;
9+
int n = prices.length;
10+
memo = new int[n][2];
11+
for (int i = 0; i < n; i++) {
12+
Arrays.fill(memo[i], -1);
13+
}
14+
return dfs(n - 1, 0);
15+
}
16+
17+
// dfs(i, 0) 表示到第 i 天结束时,未持有股票的最大利润
18+
// dfs(i, 1) 表示到第 i 天结束时,持有股票的最大利润
19+
// 第 i-1 天结束就是第 i 天的开始
20+
private int dfs(int i, int hold) {
21+
if (i < 0) {
22+
if (hold == 1) return (int) -1e9;
23+
return 0;
24+
}
25+
if (memo[i][hold] != -1) return memo[i][hold];
26+
int res;
27+
if (hold == 1) res = Math.max(dfs(i - 1, 1), dfs(i - 1, 0) - prices[i]);
28+
else res = Math.max(dfs(i - 1, 0), dfs(i - 1, 1) + prices[i]);
29+
return memo[i][hold] = res;
30+
}
31+
32+
public int maxProfit2(int[] prices) {
333
int res = 0;
434
for (int i = 0; i < prices.length - 1; i++) {
535
// 贪心。累加所有差值为正数的交易
@@ -19,6 +49,12 @@ public int maxProfit(int[] prices) {
1949
1 <= prices.length <= 3 * 10^4
2050
0 <= prices[i] <= 10^4
2151
22-
贪心。只要明天比今天涨的都进行交易。
52+
状态机 DP / 贪心。只要明天比今天涨的都进行交易。
2353
系列题解 https://leetcode.cn/problems/best-time-to-buy-and-sell-stock-ii/solution/tan-xin-suan-fa-by-liweiwei1419-2/
54+
相似题目: 309. 最佳买卖股票时机含冷冻期
55+
https://leetcode.cn/problems/best-time-to-buy-and-sell-stock-with-cooldown/
56+
714. 买卖股票的最佳时机含手续费
57+
https://leetcode.cn/problems/best-time-to-buy-and-sell-stock-with-transaction-fee/
58+
188. 买卖股票的最佳时机 IV
59+
https://leetcode.cn/problems/best-time-to-buy-and-sell-stock-iv/
2460
*/
Lines changed: 27 additions & 31 deletions
Original file line numberDiff line numberDiff line change
@@ -1,39 +1,35 @@
1-
public class Solution188 {
2-
public int maxProfit(int k, int[] prices) {
3-
int len = prices.length;
4-
if (len == 0) {
5-
return 0;
6-
}
7-
8-
// 因为 n 天最多只能进行 floor(len/2) 笔交易
9-
k = Math.min(k, len / 2);
10-
11-
// buy[i][j] 表示对于数组 prices[0,i] 中的价格而言,进行恰好 j 笔交易,并且当前手上持有一支股票,这种情况下的最大利润
12-
// sell[i][j] 表示恰好进行 j 笔交易,并且当前手上不持有股票,这种情况下的最大利润
13-
int[][] buy = new int[len][k + 1];
14-
int[][] sell = new int[len][k + 1];
1+
import java.util.Arrays;
152

16-
// 初始状态
17-
buy[0][0] = -prices[0];
18-
sell[0][0] = 0;
19-
for (int i = 1; i <= k; ++i) {
20-
buy[0][i] = sell[0][i] = Integer.MIN_VALUE / 2;
21-
}
3+
public class Solution188 {
4+
private int[] prices;
5+
private int[][][] memo;
226

23-
// 状态转移
24-
for (int i = 1; i < len; ++i) {
25-
buy[i][0] = Math.max(buy[i - 1][0], sell[i - 1][0] - prices[i]);
26-
for (int j = 1; j <= k; ++j) {
27-
buy[i][j] = Math.max(buy[i - 1][j], sell[i - 1][j] - prices[i]);
28-
sell[i][j] = Math.max(sell[i - 1][j], buy[i - 1][j - 1] + prices[i]);
7+
public int maxProfit(int k, int[] prices) {
8+
this.prices = prices;
9+
int n = prices.length;
10+
memo = new int[n][k + 1][2];
11+
for (int i = 0; i < n; i++) {
12+
for (int j = 0; j < k + 1; j++) {
13+
Arrays.fill(memo[i][j], -1);
2914
}
3015
}
16+
return dfs(n - 1, k, 0);
17+
}
3118

32-
int max = 0;
33-
for (int x : sell[len - 1]) {
34-
max = Math.max(max, x);
19+
// dfs(i, 0) 表示到第 i 天结束时完成至多 j 笔交易,未持有股票的最大利润
20+
// dfs(i, 1) 表示到第 i 天结束时完成至多 j 笔交易,持有股票的最大利润
21+
// 第 i-1 天结束就是第 i 天的开始
22+
private int dfs(int i, int j, int hold) {
23+
if (j < 0) return (int) -1e9;
24+
if (i < 0) {
25+
if (hold == 1) return (int) -1e9;
26+
return 0;
3527
}
36-
return max;
28+
if (memo[i][j][hold] != -1) return memo[i][j][hold];
29+
int res;
30+
if (hold == 1) res = Math.max(dfs(i - 1, j, 1), dfs(i - 1, j - 1, 0) - prices[i]);
31+
else res = Math.max(dfs(i - 1, j, 0), dfs(i - 1, j, 1) + prices[i]);
32+
return memo[i][j][hold] = res;
3733
}
3834
}
3935
/*
@@ -48,5 +44,5 @@ public int maxProfit(int k, int[] prices) {
4844
0 <= prices.length <= 1000
4945
0 <= prices[i] <= 1000
5046
51-
动态规划
47+
状态机 DP
5248
*/

‎leetcode/leetcode-02/src/test/java/Solution122Tests.java‎

Lines changed: 3 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -9,19 +9,22 @@ public void example1() {
99
int[] prices = {7, 1, 5, 3, 6, 4};
1010
int expected = 7;
1111
Assertions.assertEquals(expected, solution122.maxProfit(prices));
12+
Assertions.assertEquals(expected, solution122.maxProfit2(prices));
1213
}
1314

1415
@Test
1516
public void example2() {
1617
int[] prices = {1, 2, 3, 4, 5};
1718
int expected = 4;
1819
Assertions.assertEquals(expected, solution122.maxProfit(prices));
20+
Assertions.assertEquals(expected, solution122.maxProfit2(prices));
1921
}
2022

2123
@Test
2224
public void example3() {
2325
int[] prices = {7, 6, 4, 3, 1};
2426
int expected = 0;
2527
Assertions.assertEquals(expected, solution122.maxProfit(prices));
28+
Assertions.assertEquals(expected, solution122.maxProfit2(prices));
2629
}
2730
}
Lines changed: 24 additions & 30 deletions
Original file line numberDiff line numberDiff line change
@@ -1,41 +1,31 @@
1-
import java.util.ArrayDeque;
21
import java.util.ArrayList;
3-
import java.util.HashMap;
2+
import java.util.Arrays;
43
import java.util.List;
5-
import java.util.Map;
6-
import java.util.Queue;
74

85
public class Solution207 {
96
public boolean canFinish(int numCourses, int[][] prerequisites) {
10-
Map<Integer, List<Integer>> adj = new HashMap<>();
11-
int[] inDeg = new int[numCourses];
12-
for (int[] prerequisite : prerequisites) {
13-
int ai = prerequisite[1];
14-
int bi = prerequisite[0];
15-
// 其中 prerequisites[i] = [ai, bi] ,表示在选修课程 ai 前 必须 先选修 bi 。
16-
adj.computeIfAbsent(ai, key -> new ArrayList<>()).add(bi);
17-
inDeg[bi]++;
7+
List<Integer>[] g = new ArrayList[numCourses];
8+
Arrays.setAll(g, e -> new ArrayList<>());
9+
for (int[] p : prerequisites) {
10+
int x = p[1], y = p[0];
11+
g[x].add(y);
1812
}
1913

20-
// 拓扑排序
21-
Queue<Integer> queue = new ArrayDeque<>();
22-
for (int id = 0; id < numCourses; id++) {
23-
if (inDeg[id] == 0) {
24-
queue.add(id);
25-
}
14+
int[] colors = new int[numCourses];
15+
for (int i = 0; i < numCourses; i++) {
16+
if (colors[i] == 0 && dfs(i, g, colors)) return false; // 有环
2617
}
27-
List<Integer> resList = new ArrayList<>();
28-
while (!queue.isEmpty()) {
29-
int x = queue.remove();
30-
resList.add(x);
31-
for (int y : adj.getOrDefault(x, new ArrayList<>())) {
32-
inDeg[y]--;
33-
if (inDeg[y] == 0) {
34-
queue.add(y);
35-
}
36-
}
18+
return true; // 没有环
19+
}
20+
21+
private boolean dfs(int x, List<Integer>[] g, int[] colors) {
22+
colors[x] = 1; // x 正在访问中
23+
for (int y : g[x]) {
24+
if (colors[y] == 1) return true;
25+
if (colors[y] == 0 && dfs(y, g, colors)) return true;
3726
}
38-
return resList.size() == numCourses;
27+
colors[x] = 2; // x 完全访问完毕
28+
return false; // 没有找到环
3929
}
4030
}
4131
/*
@@ -54,7 +44,11 @@ public boolean canFinish(int numCourses, int[][] prerequisites) {
5444
0 <= ai, bi < numCourses
5545
prerequisites[i] 中的所有课程对 互不相同
5646
57-
拓扑排序。
47+
三色标记法 判环
48+
0 没有遍历过
49+
1 正在遍历中
50+
2 遍历过了
51+
另有 拓扑排序 解法。
5852
相似题目: 210. 课程表 II
5953
https://leetcode.cn/problems/course-schedule-ii/
6054
*/

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