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Commit 648df89

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Merge branch 'master' of github.com:youngyangyang04/leetcode-master
2 parents 1b4422d + f3ec7a5 commit 648df89

18 files changed

+331
-111
lines changed

‎README.md

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[点此这里](https://github.com/youngyangyang04/leetcode-master/graphs/contributors)查看LeetCode-Master的所有贡献者。感谢他们补充了LeetCode-Master的其他语言版本,让更多的读者收益于此项目。
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# Star 趋势
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[![Star History Chart](https://api.star-history.com/svg?repos=youngyangyang04/leetcode-master&type=Date)](https://star-history.com/#youngyangyang04/leetcode-master&Date)
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# 关于作者
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大家好,我是程序员Carl,哈工大师兄,《代码随想录》作者,先后在腾讯和百度从事后端技术研发,CSDN博客专家。对算法和C++后端技术有一定的见解,利用工作之余重新刷leetcode。

‎problems/0027.移除元素.md

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Python:
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186-
```python
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```python3
187187
class Solution:
188-
"""双指针法
189-
时间复杂度:O(n)
190-
空间复杂度:O(1)
191-
"""
192-
193-
@classmethod
194-
def removeElement(cls, nums: List[int], val: int) -> int:
195-
fast = slow = 0
196-
197-
while fast < len(nums):
198-
199-
if nums[fast] != val:
200-
nums[slow] = nums[fast]
201-
slow += 1
202-
203-
# 当 fast 指针遇到要删除的元素时停止赋值
204-
# slow 指针停止移动, fast 指针继续前进
205-
fast += 1
206-
207-
return slow
188+
def removeElement(self, nums: List[int], val: int) -> int:
189+
if nums is None or len(nums)==0:
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return 0
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l=0
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r=len(nums)-1
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while l<r:
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while(l<r and nums[l]!=val):
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l+=1
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while(l<r and nums[r]==val):
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r-=1
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nums[l], nums[r]=nums[r], nums[l]
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print(nums)
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if nums[l]==val:
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return l
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else:
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return l+1
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```
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‎problems/0028.实现strStr.md

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```
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Python3:
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```python
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//暴力解法:
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class Solution(object):
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def strStr(self, haystack, needle):
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"""
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:type haystack: str
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:type needle: str
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:rtype: int
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"""
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m,n=len(haystack),len(needle)
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for i in range(m):
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if haystack[i:i+n]==needle:
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return i
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return -1
702+
```
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```python
690704
// 方法一
691705
class Solution:

‎problems/0034.在排序数组中查找元素的第一个和最后一个位置.md

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# 34. 在排序数组中查找元素的第一个和最后一个位置
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10-
[题目链接](https://leetcode.cn/problems/find-first-and-last-position-of-element-in-sorted-array/)
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[力扣链接](https://leetcode.cn/problems/find-first-and-last-position-of-element-in-sorted-array/)
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给定一个按照升序排列的整数数组 nums,和一个目标值 target。找出给定目标值在数组中的开始位置和结束位置。
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‎problems/0046.全排列.md

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@@ -359,6 +359,36 @@ function permute(nums: number[]): number[][] {
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};
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```
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### Rust
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```Rust
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impl Solution {
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fn backtracking(result: &mut Vec<Vec<i32>>, path: &mut Vec<i32>, nums: &Vec<i32>, used: &mut Vec<bool>) {
367+
let len = nums.len();
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if path.len() == len {
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result.push(path.clone());
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return;
371+
}
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for i in 0..len {
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if used[i] == true { continue; }
374+
used[i] = true;
375+
path.push(nums[i]);
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Self::backtracking(result, path, nums, used);
377+
path.pop();
378+
used[i] = false;
379+
}
380+
}
381+
382+
pub fn permute(nums: Vec<i32>) -> Vec<Vec<i32>> {
383+
let mut result: Vec<Vec<i32>> = Vec::new();
384+
let mut path: Vec<i32> = Vec::new();
385+
let mut used = vec![false; nums.len()];
386+
Self::backtracking(&mut result, &mut path, &nums, &mut used);
387+
result
388+
}
389+
}
390+
```
391+
362392
### C
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364394
```c

‎problems/0052.N皇后II.md

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@@ -144,7 +144,61 @@ var totalNQueens = function(n) {
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};
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```
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TypeScript:
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```typescript
150+
// 0-该格为空,1-该格有皇后
151+
type GridStatus = 0 | 1;
152+
function totalNQueens(n: number): number {
153+
let resCount: number = 0;
154+
const chess: GridStatus[][] = new Array(n).fill(0)
155+
.map(_ => new Array(n).fill(0));
156+
backTracking(chess, n, 0);
157+
return resCount;
158+
function backTracking(chess: GridStatus[][], n: number, startRowIndex: number): void {
159+
if (startRowIndex === n) {
160+
resCount++;
161+
return;
162+
}
163+
for (let j = 0; j < n; j++) {
164+
if (checkValid(chess, startRowIndex, j, n) === true) {
165+
chess[startRowIndex][j] = 1;
166+
backTracking(chess, n, startRowIndex + 1);
167+
chess[startRowIndex][j] = 0;
168+
}
169+
}
170+
}
171+
};
172+
function checkValid(chess: GridStatus[][], i: number, j: number, n: number): boolean {
173+
// 向上纵向检查
174+
let tempI: number = i - 1,
175+
tempJ: number = j;
176+
while (tempI >= 0) {
177+
if (chess[tempI][tempJ] === 1) return false;
178+
tempI--;
179+
}
180+
// 斜向左上检查
181+
tempI = i - 1;
182+
tempJ = j - 1;
183+
while (tempI >= 0 && tempJ >= 0) {
184+
if (chess[tempI][tempJ] === 1) return false;
185+
tempI--;
186+
tempJ--;
187+
}
188+
// 斜向右上检查
189+
tempI = i - 1;
190+
tempJ = j + 1;
191+
while (tempI >= 0 && tempJ < n) {
192+
if (chess[tempI][tempJ] === 1) return false;
193+
tempI--;
194+
tempJ++;
195+
}
196+
return true;
197+
}
198+
```
199+
147200
C
201+
148202
```c
149203
//path[i]为在i行,path[i]列上存在皇后
150204
int *path;

‎problems/0054.螺旋矩阵.md

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return res
173173
```
174-
174+
```python3
175+
class Solution:
176+
def spiralOrder(self, matrix: List[List[int]]) -> List[int]:
177+
r=len(matrix)
178+
if r == 0 or len(matrix[0])==0:
179+
return []
180+
c=len(matrix[0])
181+
res=matrix[0]
182+
183+
if r>1:
184+
for i in range (1,r):
185+
res.append(matrix[i][c-1])
186+
for j in range(c-2, -1, -1):
187+
res.append(matrix[r-1][j])
188+
if c>1:
189+
for i in range(r-2, 0, -1):
190+
res.append(matrix[i][0])
191+
192+
M=[]
193+
for k in range(1, r-1):
194+
e=matrix[k][1:-1]
195+
M.append(e)
196+
197+
return res+self.spiralOrder(M)
198+
```
175199
-----------------------
176200
<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

‎problems/0134.加油站.md

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@@ -471,5 +471,73 @@ int canCompleteCircuit(int* gas, int gasSize, int* cost, int costSize){
471471
}
472472
```
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474+
### Scala
475+
476+
暴力解法:
477+
478+
```scala
479+
object Solution {
480+
def canCompleteCircuit(gas: Array[Int], cost: Array[Int]): Int = {
481+
for (i <- cost.indices) {
482+
var rest = gas(i) - cost(i)
483+
var index = (i + 1) % cost.length // index为i的下一个节点
484+
while (rest > 0 && i != index) {
485+
rest += (gas(index) - cost(index))
486+
index = (index + 1) % cost.length
487+
}
488+
if (rest >= 0 && index == i) return i
489+
}
490+
-1
491+
}
492+
}
493+
```
494+
495+
贪心算法,方法一:
496+
497+
```scala
498+
object Solution {
499+
def canCompleteCircuit(gas: Array[Int], cost: Array[Int]): Int = {
500+
var curSum = 0
501+
var min = Int.MaxValue
502+
for (i <- gas.indices) {
503+
var rest = gas(i) - cost(i)
504+
curSum += rest
505+
min = math.min(min, curSum)
506+
}
507+
if (curSum < 0) return -1 // 情况1: gas的总和小于cost的总和,不可能到达终点
508+
if (min >= 0) return 0 // 情况2: 最小值>=0,从0号出发可以直接到达
509+
// 情况3: min为负值,从后向前看,能把min填平的节点就是出发节点
510+
for (i <- gas.length - 1 to 0 by -1) {
511+
var rest = gas(i) - cost(i)
512+
min += rest
513+
if (min >= 0) return i
514+
}
515+
-1
516+
}
517+
}
518+
```
519+
520+
贪心算法,方法二:
521+
522+
```scala
523+
object Solution {
524+
def canCompleteCircuit(gas: Array[Int], cost: Array[Int]): Int = {
525+
var curSum = 0
526+
var totalSum = 0
527+
var start = 0
528+
for (i <- gas.indices) {
529+
curSum += (gas(i) - cost(i))
530+
totalSum += (gas(i) - cost(i))
531+
if (curSum < 0) {
532+
start = i + 1 // 起始位置更新
533+
curSum = 0 // curSum从0开始
534+
}
535+
}
536+
if (totalSum < 0) return -1 // 说明怎么走不可能跑一圈
537+
start
538+
}
539+
}
540+
```
541+
474542
-----------------------
475543
<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

‎problems/0209.长度最小的子数组.md

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179179
index += 1
180180
return 0 if res==float("inf") else res
181181
```
182-
183-
182+
```python3
183+
#滑动窗口
184+
class Solution:
185+
def minSubArrayLen(self, target: int, nums: List[int]) -> int:
186+
if nums is None or len(nums)==0:
187+
return 0
188+
lenf=len(nums)+1
189+
total=0
190+
i=j=0
191+
while (j<len(nums)):
192+
total=total+nums[j]
193+
j+=1
194+
while (total>=target):
195+
lenf=min(lenf,j-i)
196+
total=total-nums[i]
197+
i+=1
198+
if lenf==len(nums)+1:
199+
return 0
200+
else:
201+
return lenf
202+
```
184203
Go:
185204
```go
186205
func minSubArrayLen(target int, nums []int) int {

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