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Commit 07e0016

Browse files
Merge branch 'master' of github.com:youngyangyang04/leetcode-master
2 parents 59825d6 + 1923084 commit 07e0016

25 files changed

+555
-40
lines changed

‎problems/0001.两数之和.md‎

Lines changed: 9 additions & 7 deletions
Original file line numberDiff line numberDiff line change
@@ -263,13 +263,15 @@ php
263263
```php
264264
function twoSum(array $nums, int $target): array
265265
{
266-
for ($i = 0; $i < count($nums);$i++) {
267-
// 计算剩下的数
268-
$residue = $target - $nums[$i];
269-
// 匹配的index,有则返回index, 无则返回false
270-
$match_index = array_search($residue, $nums);
271-
if ($match_index !== false && $match_index != $i) {
272-
return array($i, $match_index);
266+
$map = [];
267+
foreach($nums as $i => $num) {
268+
if (isset($map[$target - $num])) {
269+
return [
270+
$i,
271+
$map[$target - $num]
272+
];
273+
} else {
274+
$map[$num] = $i;
273275
}
274276
}
275277
return [];

‎problems/0015.三数之和.md‎

Lines changed: 3 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -493,6 +493,9 @@ function threeSum(nums: number[]): number[][] {
493493
right: number = length - 1;
494494
let resArr: number[][] = [];
495495
for (let i = 0; i < length; i++) {
496+
if (nums[i]>0) {
497+
return resArr; //nums经过排序后,只要nums[i]>0, 此后的nums[i] + nums[left] + nums[right]均大于0,可以提前终止循环。
498+
}
496499
if (i > 0 && nums[i] === nums[i - 1]) {
497500
continue;
498501
}

‎problems/0019.删除链表的倒数第N个节点.md‎

Lines changed: 11 additions & 9 deletions
Original file line numberDiff line numberDiff line change
@@ -191,18 +191,20 @@ TypeScript:
191191
```typescript
192192
function removeNthFromEnd(head: ListNode | null, n: number): ListNode | null {
193193
let newHead: ListNode | null = new ListNode(0, head);
194-
let slowNode: ListNode | null = newHead,
195-
fastNode: ListNode | null = newHead;
196-
for (let i = 0; i < n; i++) {
197-
fastNode = fastNode.next;
194+
//根据leetcode题目的定义可推断这里快慢指针均不需要定义为ListNode | null。
195+
let slowNode: ListNode = newHead;
196+
let fastNode: ListNode = newHead;
197+
198+
while(n--) {
199+
fastNode = fastNode.next!; //由虚拟头节点前进n个节点时,fastNode.next可推断不为null。
198200
}
199-
while(fastNode.next) {
201+
while(fastNode.next) {//遍历直至fastNode.next = null, 即尾部节点。 此时slowNode指向倒数第n个节点。
200202
fastNode = fastNode.next;
201-
slowNode = slowNode.next;
203+
slowNode = slowNode.next!;
202204
}
203-
slowNode.next = slowNode.next.next;
204-
return newHead.next;
205-
};
205+
slowNode.next = slowNode.next!.next;//倒数第n个节点可推断其next节点不为空。
206+
return newHead.next;
207+
}
206208
```
207209

208210
版本二(计算节点总数法):

‎problems/0024.两两交换链表中的节点.md‎

Lines changed: 43 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -337,5 +337,48 @@ object Solution {
337337
}
338338
```
339339

340+
PHP:
341+
```php
342+
//虚拟头结点
343+
function swapPairs($head) {
344+
if ($head == null || $head->next == null) {
345+
return $head;
346+
}
347+
348+
$dummyNode = new ListNode(0, $head);
349+
$preNode = $dummyNode; //虚拟头结点
350+
$curNode = $head;
351+
$nextNode = $head->next;
352+
while($curNode && $nextNode) {
353+
$nextNextNode = $nextNode->next; //存下一个节点
354+
$nextNode->next = $curNode; //交换curHead 和 nextHead
355+
$curNode->next = $nextNextNode;
356+
$preNode->next = $nextNode; //上一个节点的下一个指向指向nextHead
357+
358+
//更新当前的几个指针
359+
$preNode = $preNode->next->next;
360+
$curNode = $nextNextNode;
361+
$nextNode = $nextNextNode->next;
362+
}
363+
364+
return $dummyNode->next;
365+
}
366+
367+
//递归版本
368+
function swapPairs($head)
369+
{
370+
// 终止条件
371+
if ($head === null || $head->next === null) {
372+
return $head;
373+
}
374+
375+
//结果要返回的头结点
376+
$next = $head->next;
377+
$head->next = $this->swapPairs($next->next); //当前头结点->next指向更新
378+
$next->next = $head; //当前第二个节点的->next指向更新
379+
return $next; //返回翻转后的头结点
380+
}
381+
```
382+
340383
-----------------------
341384
<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

‎problems/0056.合并区间.md‎

Lines changed: 26 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -329,5 +329,31 @@ object Solution {
329329
}
330330
```
331331

332+
### Rust
333+
334+
```Rust
335+
impl Solution {
336+
fn max(a: i32, b: i32) -> i32 {
337+
if a > b { a } else { b }
338+
}
339+
340+
pub fn merge(intervals: Vec<Vec<i32>>) -> Vec<Vec<i32>> {
341+
let mut intervals = intervals;
342+
let mut result = Vec::new();
343+
if intervals.len() == 0 { return result; }
344+
intervals.sort_by(|a, b| a[0].cmp(&b[0]));
345+
result.push(intervals[0].clone());
346+
for i in 1..intervals.len() {
347+
if result.last_mut().unwrap()[1] >= intervals[i][0] {
348+
result.last_mut().unwrap()[1] = Self::max(result.last_mut().unwrap()[1], intervals[i][1]);
349+
} else {
350+
result.push(intervals[i].clone());
351+
}
352+
}
353+
result
354+
}
355+
}
356+
```
357+
332358
-----------------------
333359
<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

‎problems/0077.组合.md‎

Lines changed: 1 addition & 1 deletion
Original file line numberDiff line numberDiff line change
@@ -114,7 +114,7 @@ vector<vector<int>> result; // 存放符合条件结果的集合
114114
vector<int> path; // 用来存放符合条件结果
115115
```
116116

117-
其实不定义这两个全局遍历也是可以的,把这两个变量放进递归函数的参数里,但函数里参数太多影响可读性,所以我定义全局变量了。
117+
其实不定义这两个全局变量也是可以的,把这两个变量放进递归函数的参数里,但函数里参数太多影响可读性,所以我定义全局变量了。
118118

119119
函数里一定有两个参数,既然是集合n里面取k的数,那么n和k是两个int型的参数。
120120

‎problems/0096.不同的二叉搜索树.md‎

Lines changed: 18 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -252,6 +252,24 @@ function numTrees(n: number): number {
252252
};
253253
```
254254

255+
### Rust
256+
257+
```Rust
258+
impl Solution {
259+
pub fn num_trees(n: i32) -> i32 {
260+
let n = n as usize;
261+
let mut dp = vec![0; n + 1];
262+
dp[0] = 1;
263+
for i in 1..=n {
264+
for j in 1..=i {
265+
dp[i] += dp[j - 1] * dp[i - j];
266+
}
267+
}
268+
dp[n]
269+
}
270+
}
271+
```
272+
255273
### C
256274

257275
```c

‎problems/0100.相同的树.md‎

Lines changed: 14 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -240,6 +240,20 @@ Go:
240240

241241
JavaScript:
242242

243+
> 递归法
244+
245+
```javascript
246+
var isSameTree = function (p, q) {
247+
if (p == null && q == null)
248+
return true;
249+
if (p == null || q == null)
250+
return false;
251+
if (p.val != q.val)
252+
return false;
253+
return isSameTree(p.left, q.left) && isSameTree(p.right, q.right);
254+
};
255+
```
256+
243257
TypeScript:
244258

245259
> 递归法-先序遍历

‎problems/0134.加油站.md‎

Lines changed: 49 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -405,6 +405,55 @@ function canCompleteCircuit(gas: number[], cost: number[]): number {
405405
};
406406
```
407407

408+
### Rust
409+
410+
贪心算法:方法一
411+
412+
```Rust
413+
impl Solution {
414+
pub fn can_complete_circuit(gas: Vec<i32>, cost: Vec<i32>) -> i32 {
415+
let mut cur_sum = 0;
416+
let mut min = i32::MAX;
417+
for i in 0..gas.len() {
418+
let rest = gas[i] - cost[i];
419+
cur_sum += rest;
420+
if cur_sum < min { min = cur_sum; }
421+
}
422+
if cur_sum < 0 { return -1; }
423+
if min > 0 { return 0; }
424+
for i in (0..gas.len()).rev() {
425+
let rest = gas[i] - cost[i];
426+
min += rest;
427+
if min >= 0 { return i as i32; }
428+
}
429+
-1
430+
}
431+
}
432+
```
433+
434+
贪心算法:方法二
435+
436+
```Rust
437+
impl Solution {
438+
pub fn can_complete_circuit(gas: Vec<i32>, cost: Vec<i32>) -> i32 {
439+
let mut cur_sum = 0;
440+
let mut total_sum = 0;
441+
let mut start = 0;
442+
for i in 0..gas.len() {
443+
cur_sum += gas[i] - cost[i];
444+
total_sum += gas[i] - cost[i];
445+
if cur_sum < 0 {
446+
start = i + 1;
447+
cur_sum = 0;
448+
}
449+
}
450+
if total_sum < 0 { return -1; }
451+
start as i32
452+
}
453+
}
454+
```
455+
456+
408457
### C
409458

410459
贪心算法:方法一

‎problems/0203.移除链表元素.md‎

Lines changed: 2 additions & 1 deletion
Original file line numberDiff line numberDiff line change
@@ -385,7 +385,8 @@ function removeElements(head: ListNode | null, val: number): ListNode | null {
385385
if (cur.val === val) {
386386
pre.next = cur.next;
387387
} else {
388-
pre = pre.next;
388+
//此处不加类型断言时:编译器会认为pre类型为ListNode, pre.next类型为ListNode | null
389+
pre = pre.next as ListNode;
389390
}
390391
cur = cur.next;
391392
}

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