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Commit c366f26

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Merge branch 'youngyangyang04:master' into master
2 parents aefb7aa + f16c640 commit c366f26

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‎README.md

Lines changed: 3 additions & 1 deletion
Original file line numberDiff line numberDiff line change
@@ -107,7 +107,9 @@
107107
5. [数组:209.长度最小的子数组](./problems/0209.长度最小的子数组.md)
108108
6. [数组:区间和](./problems/kamacoder/0058.区间和.md)
109109
6. [数组:59.螺旋矩阵II](./problems/0059.螺旋矩阵II.md)
110-
7. [数组:总结篇](./problems/数组总结篇.md)
110+
7. [数组:区间和](./problems/kamacoder/0058.区间和.md)
111+
8. [数组:开发商购买土地](./problems/kamacoder/0044.开发商购买土地.md)
112+
9. [数组:总结篇](./problems/数组总结篇.md)
111113

112114
## 链表
113115

‎problems/0018.四数之和.md

Lines changed: 1 addition & 1 deletion
Original file line numberDiff line numberDiff line change
@@ -264,7 +264,7 @@ class Solution {
264264
265265
// nums[i]+nums[j] > target 直接返回, 剪枝操作
266266
if (nums[i]+nums[j] > 0 && nums[i]+nums[j] > target) {
267-
return result;
267+
break;
268268
}
269269
270270
if (j > i + 1 && nums[j - 1] == nums[j]) { // 对nums[j]去重

‎problems/0046.全排列.md

Lines changed: 4 additions & 3 deletions
Original file line numberDiff line numberDiff line change
@@ -42,7 +42,8 @@
4242

4343
我以[1,2,3]为例,抽象成树形结构如下:
4444

45-
![46.全排列](https://code-thinking-1253855093.file.myqcloud.com/pics/20211027181706.png)
45+
46+
![全排列](https://code-thinking-1253855093.file.myqcloud.com/pics/20240803180318.png)
4647

4748
### 回溯三部曲
4849

@@ -54,7 +55,7 @@
5455

5556
但排列问题需要一个used数组,标记已经选择的元素,如图橘黄色部分所示:
5657

57-
![46.全排列](https://code-thinking-1253855093.file.myqcloud.com/pics/20211027181706.png)
58+
![全排列](https://code-thinking-1253855093.file.myqcloud.com/pics/20240803180318.png)
5859

5960
代码如下:
6061

@@ -66,7 +67,7 @@ void backtracking (vector<int>& nums, vector<bool>& used)
6667
6768
* 递归终止条件
6869
69-
![46.全排列](https://code-thinking-1253855093.file.myqcloud.com/pics/20201209174225145.png)
70+
![全排列](https://code-thinking-1253855093.file.myqcloud.com/pics/20240803180318.png)
7071
7172
可以看出叶子节点,就是收割结果的地方。
7273

‎problems/0062.不同路径.md

Lines changed: 21 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -371,6 +371,7 @@ class Solution:
371371
```
372372
### Go
373373

374+
动态规划
374375
```Go
375376
func uniquePaths(m int, n int) int {
376377
dp := make([][]int, m)
@@ -390,6 +391,26 @@ func uniquePaths(m int, n int) int {
390391
}
391392
```
392393

394+
数论方法
395+
```Go
396+
func uniquePaths(m int, n int) int {
397+
numerator := 1
398+
denominator := m - 1
399+
count := m - 1
400+
t := m + n - 2
401+
for count > 0 {
402+
numerator *= t
403+
t--
404+
for denominator != 0 && numerator % denominator == 0 {
405+
numerator /= denominator
406+
denominator--
407+
}
408+
count--
409+
}
410+
return numerator
411+
}
412+
```
413+
393414
### Javascript
394415

395416
```Javascript

‎problems/0090.子集II.md

Lines changed: 37 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -310,6 +310,43 @@ class Solution:
310310
```
311311
### Go
312312

313+
使用used数组
314+
```Go
315+
var (
316+
result [][]int
317+
path []int
318+
)
319+
320+
func subsetsWithDup(nums []int) [][]int {
321+
result = make([][]int, 0)
322+
path = make([]int, 0)
323+
used := make([]bool, len(nums))
324+
sort.Ints(nums) // 去重需要排序
325+
backtracing(nums, 0, used)
326+
return result
327+
}
328+
329+
func backtracing(nums []int, startIndex int, used []bool) {
330+
tmp := make([]int, len(path))
331+
copy(tmp, path)
332+
result = append(result, tmp)
333+
for i := startIndex; i < len(nums); i++ {
334+
// used[i - 1] == true,说明同一树枝candidates[i - 1]使用过
335+
// used[i - 1] == false,说明同一树层candidates[i - 1]使用过
336+
// 而我们要对同一树层使用过的元素进行跳过
337+
if i > 0 && nums[i] == nums[i-1] && used[i-1] == false {
338+
continue
339+
}
340+
path = append(path, nums[i])
341+
used[i] = true
342+
backtracing(nums, i + 1, used)
343+
path = path[:len(path)-1]
344+
used[i] = false
345+
}
346+
}
347+
```
348+
349+
不使用used数组
313350
```Go
314351
var (
315352
path []int

‎problems/0093.复原IP地址.md

Lines changed: 2 additions & 2 deletions
Original file line numberDiff line numberDiff line change
@@ -143,7 +143,7 @@ for (int i = startIndex; i < s.size(); i++) {
143143
代码如下:
144144

145145
```CPP
146-
// 判断字符串s在左闭又闭区间[start, end]所组成的数字是否合法
146+
// 判断字符串s在左闭右闭区间[start, end]所组成的数字是否合法
147147
bool isValid(const string& s, int start, int end) {
148148
if (start > end) {
149149
return false;
@@ -208,7 +208,7 @@ private:
208208
} else break; // 不合法,直接结束本层循环
209209
}
210210
}
211-
// 判断字符串s在左闭又闭区间[start, end]所组成的数字是否合法
211+
// 判断字符串s在左闭右闭区间[start, end]所组成的数字是否合法
212212
bool isValid(const string& s, int start, int end) {
213213
if (start > end) {
214214
return false;

‎problems/0110.平衡二叉树.md

Lines changed: 60 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -623,6 +623,8 @@ class Solution:
623623
```
624624
### Go:
625625

626+
递归法
627+
626628
```Go
627629
func isBalanced(root *TreeNode) bool {
628630
h := getHeight(root)
@@ -653,6 +655,64 @@ func max(a, b int) int {
653655
}
654656
```
655657

658+
迭代法
659+
660+
```Go
661+
func isBalanced(root *TreeNode) bool {
662+
st := make([]*TreeNode, 0)
663+
if root == nil {
664+
return true
665+
}
666+
st = append(st, root)
667+
for len(st) > 0 {
668+
node := st[len(st)-1]
669+
st = st[:len(st)-1]
670+
if math.Abs(float64(getDepth(node.Left)) - float64(getDepth(node.Right))) > 1 {
671+
return false
672+
}
673+
if node.Right != nil {
674+
st = append(st, node.Right)
675+
}
676+
if node.Left != nil {
677+
st = append(st, node.Left)
678+
}
679+
}
680+
return true
681+
}
682+
683+
func getDepth(cur *TreeNode) int {
684+
st := make([]*TreeNode, 0)
685+
if cur != nil {
686+
st = append(st, cur)
687+
}
688+
depth := 0
689+
result := 0
690+
for len(st) > 0 {
691+
node := st[len(st)-1]
692+
if node != nil {
693+
st = st[:len(st)-1]
694+
st = append(st, node, nil)
695+
depth++
696+
if node.Right != nil {
697+
st = append(st, node.Right)
698+
}
699+
if node.Left != nil {
700+
st = append(st, node.Left)
701+
}
702+
} else {
703+
st = st[:len(st)-1]
704+
node = st[len(st)-1]
705+
st = st[:len(st)-1]
706+
depth--
707+
}
708+
if result < depth {
709+
result = depth
710+
}
711+
}
712+
return result
713+
}
714+
```
715+
656716
### JavaScript:
657717

658718
递归法:

‎problems/0112.路径总和.md

Lines changed: 42 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -727,6 +727,48 @@ class Solution:
727727

728728
```go
729729
//递归法
730+
/**
731+
* Definition for a binary tree node.
732+
* type TreeNode struct {
733+
* Val int
734+
* Left *TreeNode
735+
* Right *TreeNode
736+
* }
737+
*/
738+
func hasPathSum(root *TreeNode, targetSum int) bool {
739+
if root == nil {
740+
return false
741+
}
742+
return traversal(root, targetSum - root.Val)
743+
}
744+
745+
func traversal(cur *TreeNode, count int) bool {
746+
if cur.Left == nil && cur.Right == nil && count == 0 {
747+
return true
748+
}
749+
if cur.Left == nil && cur.Right == nil {
750+
return false
751+
}
752+
if cur.Left != nil {
753+
count -= cur.Left.Val
754+
if traversal(cur.Left, count) {
755+
return true
756+
}
757+
count += cur.Left.Val
758+
}
759+
if cur.Right != nil {
760+
count -= cur.Right.Val
761+
if traversal(cur.Right, count) {
762+
return true
763+
}
764+
count += cur.Right.Val
765+
}
766+
return false
767+
}
768+
```
769+
770+
```go
771+
//递归法精简
730772
/**
731773
* Definition for a binary tree node.
732774
* type TreeNode struct {

‎problems/0121.买卖股票的最佳时机.md

Lines changed: 1 addition & 1 deletion
Original file line numberDiff line numberDiff line change
@@ -308,7 +308,7 @@ class Solution:
308308
class Solution:
309309
def maxProfit(self, prices: List[int]) -> int:
310310
length = len(prices)
311-
if len == 0:
311+
if length == 0:
312312
return 0
313313
dp = [[0] * 2 for _ in range(length)]
314314
dp[0][0] = -prices[0]

‎problems/0131.分割回文串.md

Lines changed: 58 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -527,6 +527,7 @@ class Solution:
527527

528528
```
529529
### Go
530+
回溯 基本版
530531
```go
531532
var (
532533
path []string // 放已经回文的子串
@@ -565,6 +566,63 @@ func isPalindrome(s string) bool {
565566
}
566567
```
567568

569+
回溯+动态规划优化回文串判断
570+
```go
571+
var (
572+
result [][]string
573+
path []string // 放已经回文的子串
574+
isPalindrome [][]bool // 放事先计算好的是否回文子串的结果
575+
)
576+
577+
func partition(s string) [][]string {
578+
result = make([][]string, 0)
579+
path = make([]string, 0)
580+
computePalindrome(s)
581+
backtracing(s, 0)
582+
return result
583+
}
584+
585+
func backtracing(s string, startIndex int) {
586+
// 如果起始位置已经大于s的大小,说明已经找到了一组分割方案了
587+
if startIndex >= len(s) {
588+
tmp := make([]string, len(path))
589+
copy(tmp, path)
590+
result = append(result, tmp)
591+
return
592+
}
593+
for i := startIndex; i < len(s); i++ {
594+
if isPalindrome[startIndex][i] { // 是回文子串
595+
// 获取[startIndex,i]在s中的子串
596+
path = append(path, s[startIndex:i+1])
597+
} else { // 不是回文,跳过
598+
continue
599+
}
600+
backtracing(s, i + 1) // 寻找i+1为起始位置的子串
601+
path = path[:len(path)-1] // 回溯过程,弹出本次已经添加的子串
602+
}
603+
}
604+
605+
func computePalindrome(s string) {
606+
// isPalindrome[i][j] 代表 s[i:j](双边包括)是否是回文字串
607+
isPalindrome = make([][]bool, len(s))
608+
for i := 0; i < len(isPalindrome); i++ {
609+
isPalindrome[i] = make([]bool, len(s))
610+
}
611+
for i := len(s)-1; i >= 0; i-- {
612+
// 需要倒序计算, 保证在i行时, i+1行已经计算好了
613+
for j := i; j < len(s); j++ {
614+
if j == i {
615+
isPalindrome[i][j] = true
616+
} else if j - i == 1 {
617+
isPalindrome[i][j] = s[i] == s[j]
618+
} else {
619+
isPalindrome[i][j] = s[i] == s[j] && isPalindrome[i+1][j-1]
620+
}
621+
}
622+
}
623+
}
624+
```
625+
568626
### JavaScript
569627

570628
```js

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