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Commit 57cec75

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Merge branch 'youngyangyang04:master' into master
2 parents a4f9f01 + 90fece3 commit 57cec75

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‎.DS_Store

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‎.gitignore

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.idea/
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.DS_Store
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.vscode
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.temp
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.cache
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*.iml
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__pycache__

‎README.md

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24. [图论:Bellman_ford 算法](./problems/kamacoder/0094.城市间货物运输I.md)
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25. [图论:Bellman_ford 队列优化算法(又名SPFA)](./problems/kamacoder/0094.城市间货物运输I-SPFA.md)
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26. [图论:Bellman_ford之判断负权回路](./problems/kamacoder/0095.城市间货物运输II.md)
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27. [图论:Bellman_ford之单源有限最短路](./problems/kamacoder/0095.城市间货物运输II.md)
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27. [图论:Bellman_ford之单源有限最短路](./problems/kamacoder/0096.城市间货物运输III.md)
404404
28. [图论:Floyd 算法](./problems/kamacoder/0097.小明逛公园.md)
405405
29. [图论:A * 算法](./problems/kamacoder/0126.骑士的攻击astar.md)
406406
30. [图论:最短路算法总结篇](./problems/kamacoder/最短路问题总结篇.md)

‎problems/0005.最长回文子串.md

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* 时间复杂度:O(n^2)
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* 空间复杂度:O(1)
258258

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### Manacher 算法
259260

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Manacher 算法的关键在于高效利用回文的对称性,通过插入分隔符和维护中心、边界等信息,在线性时间内找到最长回文子串。这种方法避免了重复计算,是处理回文问题的最优解。
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263+
```c++
264+
//Manacher 算法
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class Solution {
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public:
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string longestPalindrome(string s) {
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// 预处理字符串,在每个字符之间插入 '#'
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string t = "#";
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for (char c : s) {
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t += c; // 添加字符
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t += '#';// 添加分隔符
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}
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int n = t.size();// 新字符串的长度
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vector<int> p(n, 0);// p[i] 表示以 t[i] 为中心的回文半径
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int center = 0, right = 0;// 当前回文的中心和右边界
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// 遍历预处理后的字符串
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for (int i = 0; i < n; i++) {
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// 如果当前索引在右边界内,利用对称性初始化 p[i]
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if (i < right) {
283+
p[i] = min(right - i, p[2 * center - i]);
284+
}
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// 尝试扩展回文
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while (i - p[i] - 1 >= 0 && i + p[i] + 1 < n && t[i - p[i] - 1] == t[i + p[i] + 1]) {
287+
p[i]++;// 增加回文半径
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}
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// 如果当前回文扩展超出右边界,更新中心和右边界
290+
if (i + p[i] > right) {
291+
center = i;// 更新中心
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right = i + p[i];// 更新右边界
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}
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}
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// 找到最大回文半径和对应的中心
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int maxLen = 0, centerIndex = 0;
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for (int i = 0; i < n; i++) {
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if (p[i] > maxLen) {
299+
maxLen = p[i];// 更新最大回文长度
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centerIndex = i;// 更新中心索引
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}
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}
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// 计算原字符串中回文子串的起始位置并返回
304+
return s.substr((centerIndex - maxLen) / 2, maxLen);
305+
}
306+
};
307+
```
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309+
310+
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* 时间复杂度:O(n)
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* 空间复杂度:O(n)
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## 其他语言版本
262315

@@ -682,3 +735,4 @@ public class Solution {
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<a href="https://programmercarl.com/other/kstar.html" target="_blank">
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<img src="../pics/网站星球宣传海报.jpg" width="1000"/>
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</a>
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‎problems/0046.全排列.md

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public void backtrack(int[] nums, LinkedList<Integer> path) {
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if (path.size() == nums.length) {
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result.add(new ArrayList<>(path));
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return;
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}
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for (int i =0; i < nums.length; i++) {
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// 如果path中已有,则跳过
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<a href="https://programmercarl.com/other/kstar.html" target="_blank">
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<img src="../pics/网站星球宣传海报.jpg" width="1000"/>
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</a>
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‎problems/0053.最大子序和.md

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return result
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```
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贪心法
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```python
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class Solution:
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def maxSubArray(self, nums):
@@ -226,9 +227,55 @@ class Solution:
226227
if count <= 0: # 相当于重置最大子序起始位置,因为遇到负数一定是拉低总和
227228
count = 0
228229
return result
230+
```
231+
动态规划
232+
```python
233+
class Solution:
234+
def maxSubArray(self, nums: List[int]) -> int:
235+
dp = [0] * len(nums)
236+
dp[0] = nums[0]
237+
res = nums[0]
238+
for i in range(1, len(nums)):
239+
dp[i] = max(dp[i-1] + nums[i], nums[i])
240+
res = max(res, dp[i])
241+
return res
242+
```
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244+
动态规划
245+
246+
```python
247+
class Solution:
248+
def maxSubArray(self, nums):
249+
if not nums:
250+
return 0
251+
dp = [0] * len(nums) # dp[i]表示包括i之前的最大连续子序列和
252+
dp[0] = nums[0]
253+
result = dp[0]
254+
for i in range(1, len(nums)):
255+
dp[i] = max(dp[i-1]+nums[i], nums[i]) # 状态转移公式
256+
if dp[i] > result:
257+
result = dp[i] # result 保存dp[i]的最大值
258+
return result
259+
```
260+
261+
动态规划优化
229262

263+
```python
264+
class Solution:
265+
def maxSubArray(self, nums: List[int]) -> int:
266+
max_sum = float("-inf") # 初始化结果为负无穷大,方便比较取最大值
267+
current_sum = 0 # 初始化当前连续和
268+
269+
for num in nums:
230270

271+
# 更新当前连续和
272+
# 如果原本的连续和加上当前数字之后没有当前数字大,说明原本的连续和是负数,那么就直接从当前数字开始重新计算连续和
273+
current_sum = max(current_sum+num, num)
274+
max_sum = max(max_sum, current_sum) # 更新结果
275+
276+
return max_sum
231277
```
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### Go
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贪心法
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```go

‎problems/0055.跳跃游戏.md

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return False
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```
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```python
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## 基于当前最远可到达位置判断
148+
class Solution:
149+
def canJump(self, nums: List[int]) -> bool:
150+
far = nums[0]
151+
for i in range(len(nums)):
152+
# 要考虑两个情况
153+
# 1. i <= far - 表示 当前位置i 可以到达
154+
# 2. i > far - 表示 当前位置i 无法到达
155+
if i > far:
156+
return False
157+
far = max(far, nums[i]+i)
158+
# 如果循环正常结束,表示最后一个位置也可以到达,否则会在中途直接退出
159+
# 关键点在于,要想明白其实列表中的每个位置都是需要验证能否到达的
160+
return True
161+
```
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### Go
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```go

‎problems/0070.爬楼梯完全背包版本.md

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184184

185185
### Go:
186186
```go
187+
package main
188+
189+
import (
190+
"bufio"
191+
"fmt"
192+
"os"
193+
"strconv"
194+
"strings"
195+
)
196+
187197
func climbStairs(n int, m int) int {
188198
dp := make([]int, n+1)
189199
dp[0] = 1

‎problems/0093.复原IP地址.md

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num = int(s[start:end+1])
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return 0 <= num <= 255
469469

470+
回溯(版本三)
470471

471-
472-
472+
```python
473+
class Solution:
474+
def restoreIpAddresses(self, s: str) -> List[str]:
475+
result = []
476+
self.backtracking(s, 0, [], result)
477+
return result
478+
479+
def backtracking(self, s, startIndex, path, result):
480+
if startIndex == len(s):
481+
result.append('.'.join(path[:]))
482+
return
483+
484+
for i in range(startIndex, min(startIndex+3, len(s))):
485+
# 如果 i 往后遍历了,并且当前地址的第一个元素是 0 ,就直接退出
486+
if i > startIndex and s[startIndex] == '0':
487+
break
488+
# 比如 s 长度为 5,当前遍历到 i = 3 这个元素
489+
# 因为还没有执行任何操作,所以此时剩下的元素数量就是 5 - 3 = 2 ,即包括当前的 i 本身
490+
# path 里面是当前包含的子串,所以有几个元素就表示储存了几个地址
491+
# 所以 (4 - len(path)) * 3 表示当前路径至多能存放的元素个数
492+
# 4 - len(path) 表示至少要存放的元素个数
493+
if (4 - len(path)) * 3 < len(s) - i or 4 - len(path) > len(s) - i:
494+
break
495+
if i - startIndex == 2:
496+
if not int(s[startIndex:i+1]) <= 255:
497+
break
498+
path.append(s[startIndex:i+1])
499+
self.backtracking(s, i+1, path, result)
500+
path.pop()
473501
```
474502

475503
### Go

‎problems/0110.平衡二叉树.md

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while stack:
610610
node = stack.pop()
611611
if node:
612-
stack.append(node)
612+
stack.append(node)# 中
613613
stack.append(None)
614-
if node.left: stack.append(node.left)
615-
if node.right: stack.append(node.right)
614+
# 采用数组进行迭代,先将右节点加入,保证左节点能够先出栈
615+
if node.right: # 右
616+
stack.append(node.right)
617+
if node.left: # 左
618+
stack.append(node.left)
616619
else:
617620
real_node = stack.pop()
618621
left, right = height_map.get(real_node.left, 0), height_map.get(real_node.right, 0)

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