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Commit be3f885

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Merge branch 'youngyangyang04:master' into master
2 parents 7ac6785 + 110823c commit be3f885

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4 files changed

+169
-24
lines changed

4 files changed

+169
-24
lines changed

‎problems/0200.岛屿数量.深搜版.md‎

Lines changed: 1 addition & 1 deletion
Original file line numberDiff line numberDiff line change
@@ -37,7 +37,7 @@
3737

3838
在遇到标记过的陆地节点和海洋节点的时候直接跳过。 这样计数器就是最终岛屿的数量。
3939

40-
那么如果把节点陆地所能遍历到的陆地都标记上呢,就可以使用 DFS,BFS或者并查集。
40+
那么如何把节点陆地所能遍历到的陆地都标记上呢,就可以使用 DFS,BFS或者并查集。
4141

4242
### 深度优先搜索
4343

‎problems/0647.回文子串.md‎

Lines changed: 42 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -397,6 +397,7 @@ class Solution:
397397
```
398398
399399
### Go:
400+
> 动态规划:
400401
401402
```Go
402403
func countSubstrings(s string) int {
@@ -422,6 +423,47 @@ func countSubstrings(s string) int {
422423
return res
423424
}
424425
```
426+
> 动态规划:简洁版
427+
```Go
428+
func countSubstrings(s string) int {
429+
res := 0
430+
dp := make([][]bool, len(s))
431+
for i := 0; i < len(s); i++ {
432+
dp[i] = make([]bool, len(s))
433+
}
434+
435+
for i := len(s) - 1; i >= 0; i-- {
436+
for j := i; j < len(s); j++ {
437+
if s[i] == s[j] && (j-i <= 1 || dp[i+1][j-1]) {
438+
res++
439+
dp[i][j] = true
440+
}
441+
}
442+
}
443+
return res
444+
}
445+
```
446+
447+
> 双指针法:
448+
```Go
449+
func countSubstrings(s string) int {
450+
extend := func(i, j int) int {
451+
res := 0
452+
for i >= 0 && j < len(s) && s[i] == s[j] {
453+
i --
454+
j ++
455+
res ++
456+
}
457+
return res
458+
}
459+
res := 0
460+
for i := 0; i < len(s); i++ {
461+
res += extend(i, i) // 以i为中心
462+
res += extend(i, i+1) // 以i和i+1为中心
463+
}
464+
return res
465+
}
466+
```
425467

426468
### Javascript:
427469

‎problems/0827.最大人工岛.md‎

Lines changed: 93 additions & 2 deletions
Original file line numberDiff line numberDiff line change
@@ -348,6 +348,99 @@ class Solution:
348348

349349
```
350350

351+
352+
### Go
353+
354+
```go
355+
func largestIsland(grid [][]int) int {
356+
dir := [][]int{{0, 1}, {1, 0}, {-1, 0}, {0, -1}}
357+
n := len(grid)
358+
m := len(grid[0])
359+
area := 0
360+
visited := make([][]bool, n)
361+
for i := 0; i < n; i++ {
362+
visited[i] = make([]bool, m)
363+
}
364+
gridNum := make(map[int]int, 0) // 记录每一个岛屿的面积
365+
mark := 2 // 记录每个岛屿的编号
366+
isAllGrid := true
367+
res := 0 // 标记是否整个地图都是陆地
368+
369+
var dfs func(grid [][]int, visited [][]bool, x, y, mark int)
370+
dfs = func(grid [][]int, visited [][]bool, x, y, mark int) {
371+
// 终止条件:访问过的节点 或者 遇到海水
372+
if visited[x][y] || grid[x][y] == 0 {
373+
return
374+
}
375+
visited[x][y] = true // 标记访问过
376+
grid[x][y] = mark // 给陆地标记新标签
377+
area++
378+
for i := 0; i < 4; i++ {
379+
nextX := x + dir[i][0]
380+
nextY := y + dir[i][1]
381+
if nextX < 0 || nextX >= len(grid) || nextY < 0 || nextY >= len(grid[0]) {
382+
continue
383+
}
384+
dfs(grid, visited, nextX, nextY, mark)
385+
}
386+
}
387+
388+
for i := 0; i < n; i++ {
389+
for j := 0; j < m; j++ {
390+
if grid[i][j] == 0 {
391+
isAllGrid = false
392+
}
393+
if !visited[i][j] && grid[i][j] == 1 {
394+
area = 0
395+
dfs(grid, visited, i, j, mark) // 将与其链接的陆地都标记上 true
396+
gridNum[mark] = area // 记录每一个岛屿的面积
397+
mark++ // 更新下一个岛屿编号
398+
}
399+
}
400+
}
401+
if isAllGrid {
402+
return n * m
403+
}
404+
// 根据添加陆地的位置,计算周边岛屿面积之和
405+
visitedGrid := make(map[int]struct{}) // 标记访问过的岛屿
406+
for i := 0; i < n; i++ {
407+
for j := 0; j < m; j++ {
408+
count := 1 // 记录连接之后的岛屿数量
409+
visitedGrid = make(map[int]struct{}) // 每次使用时,清空
410+
if grid[i][j] == 0 {
411+
for k := 0; k < 4; k++ {
412+
// 计算相邻坐标
413+
nearI := i + dir[k][0]
414+
nearJ := j + dir[k][1]
415+
if nearI < 0 || nearI >= len(grid) || nearJ < 0 || nearJ >= len(grid[0]) {
416+
continue
417+
}
418+
// 添加过的岛屿不要重复添加
419+
if _, ok := visitedGrid[grid[nearI][nearJ]]; ok {
420+
continue
421+
}
422+
// 把相邻四面的岛屿数量加起来
423+
count += gridNum[grid[nearI][nearJ]]
424+
// 标记该岛屿已经添加过
425+
visitedGrid[grid[nearI][nearJ]] = struct{}{}
426+
}
427+
}
428+
res = max827(res, count)
429+
}
430+
}
431+
return res
432+
}
433+
434+
func max827(x, y int) int {
435+
if x > y {
436+
return x
437+
}
438+
return y
439+
}
440+
441+
```
442+
443+
351444
### JavaScript
352445

353446
```JavaScript
@@ -411,8 +504,6 @@ return res;
411504
```
412505

413506

414-
415-
416507
<p align="center">
417508
<a href="https://programmercarl.com/other/kstar.html" target="_blank">
418509
<img src="../pics/网站星球宣传海报.jpg" width="1000"/>

‎problems/二叉树的递归遍历.md‎

Lines changed: 33 additions & 21 deletions
Original file line numberDiff line numberDiff line change
@@ -180,40 +180,52 @@ class Solution {
180180

181181
class Solution:
182182
def preorderTraversal(self, root: TreeNode) -> List[int]:
183-
if not root:
184-
return []
185-
186-
left = self.preorderTraversal(root.left)
187-
right = self.preorderTraversal(root.right)
188-
189-
return [root.val] + left + right
183+
res = []
184+
185+
def dfs(node):
186+
if node is None:
187+
return
188+
189+
res.append(node.val)
190+
dfs(node.left)
191+
dfs(node.right)
192+
193+
return res
190194

191195
```
192196
```python
193197
# 中序遍历-递归-LC94_二叉树的中序遍历
194198
class Solution:
195199
def inorderTraversal(self, root: TreeNode) -> List[int]:
196-
if root is None:
197-
return []
198-
199-
left = self.inorderTraversal(root.left)
200-
right = self.inorderTraversal(root.right)
201-
202-
return left + [root.val] + right
200+
res = []
201+
202+
def dfs(node):
203+
if node is None:
204+
return
205+
206+
dfs(node.left)
207+
res.append(node.val)
208+
dfs(node.right)
209+
210+
return res
203211
```
204212
```python
205213

206214

207215
# 后序遍历-递归-LC145_二叉树的后序遍历
208216
class Solution:
209217
def postorderTraversal(self, root: TreeNode) -> List[int]:
210-
if not root:
211-
return []
212-
213-
left = self.postorderTraversal(root.left)
214-
right = self.postorderTraversal(root.right)
215-
216-
return left + right + [root.val]
218+
res = []
219+
220+
def dfs(node):
221+
if node is None:
222+
return
223+
224+
dfs(node.left)
225+
dfs(node.right)
226+
res.append(node.val)
227+
228+
return res
217229
```
218230

219231
### Go:

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