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Commit 8c2737d

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Merge branch 'master' into patch08
2 parents 16c6abf + f03f8d2 commit 8c2737d

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‎README.md‎

Lines changed: 1 addition & 1 deletion
Original file line numberDiff line numberDiff line change
@@ -254,7 +254,7 @@
254254
33. [二叉树:构造一棵搜索树](./problems/0108.将有序数组转换为二叉搜索树.md)
255255
34. [二叉树:搜索树转成累加树](./problems/0538.把二叉搜索树转换为累加树.md)
256256
35. [二叉树:总结篇!(需要掌握的二叉树技能都在这里了)](./problems/二叉树总结篇.md)
257-
257+
258258
## 回溯算法
259259

260260
题目分类大纲如下:

‎problems/0001.两数之和.md‎

Lines changed: 22 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -274,6 +274,7 @@ class Solution {
274274
}
275275
}
276276
```
277+
277278
Scala:
278279
```scala
279280
object Solution {
@@ -296,5 +297,26 @@ object Solution {
296297
}
297298
}
298299
```
300+
301+
C#:
302+
```csharp
303+
public class Solution {
304+
public int[] TwoSum(int[] nums, int target) {
305+
Dictionary<int ,int> dic= new Dictionary<int,int>();
306+
for(int i=0;i<nums.Length;i++){
307+
int imp= target-nums[i];
308+
if(dic.ContainsKey(imp)&&dic[imp]!=i){
309+
return new int[]{i, dic[imp]};
310+
}
311+
if(!dic.ContainsKey(nums[i])){
312+
dic.Add(nums[i],i);
313+
}
314+
}
315+
return new int[]{0, 0};
316+
}
317+
}
318+
```
319+
320+
299321
-----------------------
300322
<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

‎problems/0015.三数之和.md‎

Lines changed: 26 additions & 44 deletions
Original file line numberDiff line numberDiff line change
@@ -313,54 +313,36 @@ func threeSum(nums []int)[][]int{
313313
javaScript:
314314

315315
```js
316-
/**
317-
* @param {number[]} nums
318-
* @return {number[][]}
319-
*/
320-
321-
// 循环内不考虑去重
322-
var threeSum = function(nums) {
323-
const len = nums.length;
324-
if(len < 3) return [];
325-
nums.sort((a, b) => a - b);
326-
const resSet = new Set();
327-
for(let i = 0; i < len - 2; i++) {
328-
if(nums[i] > 0) break;
329-
let l = i + 1, r = len - 1;
330-
while(l < r) {
331-
const sum = nums[i] + nums[l] + nums[r];
332-
if(sum < 0) { l++; continue };
333-
if(sum > 0) { r--; continue };
334-
resSet.add(`${nums[i]},${nums[l]},${nums[r]}`);
335-
l++;
336-
r--;
337-
}
338-
}
339-
return Array.from(resSet).map(i => i.split(","));
340-
};
341-
342-
// 去重优化
343316
var threeSum = function(nums) {
344-
const len = nums.length;
345-
if(len <3) return [];
346-
nums.sort((a, b) => a - b);
347-
constres= [];
348-
for(let i = 0; i <len - 2; i++) {
349-
if(nums[i] >0) break;
350-
// a去重
351-
if(i >0&& nums[i] === nums[i -1]) continue;
352-
let l = i +1, r = len - 1;
317+
const res= [], len = nums.length
318+
// 将数组排序
319+
nums.sort((a, b) => a - b)
320+
for (let i =0; i < len; i++) {
321+
let l = i +1, r =len - 1, iNum = nums[i]
322+
// 数组排过序,如果第一个数大于0直接返回res
323+
if (iNum >0) return res
324+
// 去重
325+
if (iNum == nums[i - 1]) continue
353326
while(l < r) {
354-
const sum = nums[i] + nums[l] + nums[r];
355-
if(sum < 0) { l++; continue };
356-
if(sum > 0) { r--; continue };
357-
res.push([nums[i], nums[l], nums[r]])
358-
// b c 去重
359-
while(l < r && nums[l] === nums[++l]);
360-
while(l < r && nums[r] === nums[--r]);
327+
let lNum = nums[l], rNum = nums[r], threeSum = iNum + lNum + rNum
328+
// 三数之和小于0,则左指针向右移动
329+
if (threeSum < 0) l++
330+
else if (threeSum > 0) r--
331+
else {
332+
res.push([iNum, lNum, rNum])
333+
// 去重
334+
while(l < r && nums[l] == nums[l + 1]){
335+
l++
336+
}
337+
while(l < r && nums[r] == nums[r - 1]) {
338+
r--
339+
}
340+
l++
341+
r--
342+
}
361343
}
362344
}
363-
return res;
345+
return res
364346
};
365347
```
366348
TypeScript:

‎problems/0019.删除链表的倒数第N个节点.md‎

Lines changed: 24 additions & 2 deletions
Original file line numberDiff line numberDiff line change
@@ -39,7 +39,7 @@
3939

4040
分为如下几步:
4141

42-
* 首先这里我推荐大家使用虚拟头结点,这样方面处理删除实际头结点的逻辑,如果虚拟头结点不清楚,可以看这篇: [链表:听说用虚拟头节点会方便很多?](https://programmercarl.com/0203.移除链表元素.html)
42+
* 首先这里我推荐大家使用虚拟头结点,这样方便处理删除实际头结点的逻辑,如果虚拟头结点不清楚,可以看这篇: [链表:听说用虚拟头节点会方便很多?](https://programmercarl.com/0203.移除链表元素.html)
4343

4444
* 定义fast指针和slow指针,初始值为虚拟头结点,如图:
4545

@@ -289,6 +289,28 @@ func removeNthFromEnd(_ head: ListNode?, _ n: Int) -> ListNode? {
289289
return dummyHead.next
290290
}
291291
```
292-
292+
Scala:
293+
```scala
294+
object Solution {
295+
def removeNthFromEnd(head: ListNode, n: Int): ListNode = {
296+
val dummy = new ListNode(-1, head) // 定义虚拟头节点
297+
var fast = head // 快指针从头开始走
298+
var slow = dummy // 慢指针从虚拟头开始头
299+
// 因为参数 n 是不可变量,所以不能使用 while(n>0){n-=1}的方式
300+
for (i <- 0 until n) {
301+
fast = fast.next
302+
}
303+
// 快指针和满指针一起走,直到fast走到null
304+
while (fast != null) {
305+
slow = slow.next
306+
fast = fast.next
307+
}
308+
// 删除slow的下一个节点
309+
slow.next = slow.next.next
310+
// 返回虚拟头节点的下一个
311+
dummy.next
312+
}
313+
}
314+
```
293315
-----------------------
294316
<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

‎problems/0024.两两交换链表中的节点.md‎

Lines changed: 23 additions & 1 deletion
Original file line numberDiff line numberDiff line change
@@ -311,7 +311,29 @@ func swapPairs(_ head: ListNode?) -> ListNode? {
311311
return dummyHead.next
312312
}
313313
```
314-
314+
Scala:
315+
```scala
316+
// 虚拟头节点
317+
object Solution {
318+
def swapPairs(head: ListNode): ListNode = {
319+
var dummy = new ListNode(0, head) // 虚拟头节点
320+
var pre = dummy
321+
var cur = head
322+
// 当pre的下一个和下下个都不为空,才进行两两转换
323+
while (pre.next != null && pre.next.next != null) {
324+
var tmp: ListNode = cur.next.next // 缓存下一次要进行转换的第一个节点
325+
pre.next = cur.next // 步骤一
326+
cur.next.next = cur // 步骤二
327+
cur.next = tmp // 步骤三
328+
// 下面是准备下一轮的交换
329+
pre = cur
330+
cur = tmp
331+
}
332+
// 最终返回dummy虚拟头节点的下一个,return可以省略
333+
dummy.next
334+
}
335+
}
336+
```
315337

316338
-----------------------
317339
<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

‎problems/0027.移除元素.md‎

Lines changed: 16 additions & 2 deletions
Original file line numberDiff line numberDiff line change
@@ -81,7 +81,7 @@ public:
8181

8282
**双指针法(快慢指针法)在数组和链表的操作中是非常常见的,很多考察数组、链表、字符串等操作的面试题,都使用双指针法。**
8383

84-
后序都会一一介绍到,本题代码如下:
84+
后续都会一一介绍到,本题代码如下:
8585

8686
```CPP
8787
// 时间复杂度:O(n)
@@ -328,6 +328,20 @@ int removeElement(int* nums, int numsSize, int val){
328328
return slow;
329329
}
330330
```
331-
331+
Scala:
332+
```scala
333+
object Solution {
334+
def removeElement(nums: Array[Int], `val`: Int): Int = {
335+
var slow = 0
336+
for (fast <- 0 until nums.length) {
337+
if (`val` != nums(fast)) {
338+
nums(slow) = nums(fast)
339+
slow += 1
340+
}
341+
}
342+
slow
343+
}
344+
}
345+
```
332346
-----------------------
333347
<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

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

Lines changed: 46 additions & 1 deletion
Original file line numberDiff line numberDiff line change
@@ -480,7 +480,52 @@ var searchRange = function(nums, target) {
480480
return [-1, -1];
481481
};
482482
```
483-
483+
### Scala
484+
```scala
485+
object Solution {
486+
def searchRange(nums: Array[Int], target: Int): Array[Int] = {
487+
var left = getLeftBorder(nums, target)
488+
var right = getRightBorder(nums, target)
489+
if (left == -2 || right == -2) return Array(-1, -1)
490+
if (right - left > 1) return Array(left + 1, right - 1)
491+
Array(-1, -1)
492+
}
493+
494+
// 寻找左边界
495+
def getLeftBorder(nums: Array[Int], target: Int): Int = {
496+
var leftBorder = -2
497+
var left = 0
498+
var right = nums.length - 1
499+
while (left <= right) {
500+
var mid = left + (right - left) / 2
501+
if (nums(mid) >= target) {
502+
right = mid - 1
503+
leftBorder = right
504+
} else {
505+
left = mid + 1
506+
}
507+
}
508+
leftBorder
509+
}
510+
511+
// 寻找右边界
512+
def getRightBorder(nums: Array[Int], target: Int): Int = {
513+
var rightBorder = -2
514+
var left = 0
515+
var right = nums.length - 1
516+
while (left <= right) {
517+
var mid = left + (right - left) / 2
518+
if (nums(mid) <= target) {
519+
left = mid + 1
520+
rightBorder = left
521+
} else {
522+
right = mid - 1
523+
}
524+
}
525+
rightBorder
526+
}
527+
}
528+
```
484529

485530
-----------------------
486531
<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

‎problems/0035.搜索插入位置.md‎

Lines changed: 45 additions & 1 deletion
Original file line numberDiff line numberDiff line change
@@ -316,8 +316,52 @@ func searchInsert(_ nums: [Int], _ target: Int) -> Int {
316316
return right + 1
317317
}
318318
```
319+
### Scala
320+
```scala
321+
object Solution {
322+
def searchInsert(nums: Array[Int], target: Int): Int = {
323+
var left = 0
324+
var right = nums.length - 1
325+
while (left <= right) {
326+
var mid = left + (right - left) / 2
327+
if (target == nums(mid)) {
328+
return mid
329+
} else if (target > nums(mid)) {
330+
left = mid + 1
331+
} else {
332+
right = mid - 1
333+
}
334+
}
335+
right + 1
336+
}
337+
}
338+
```
319339

320-
340+
### PHP
341+
342+
```php
343+
// 二分法(1):[左闭右闭]
344+
function searchInsert($nums, $target)
345+
{
346+
$n = count($nums);
347+
$l = 0;
348+
$r = $n - 1;
349+
while ($l <= $r) {
350+
$mid = floor(($l + $r) / 2);
351+
if ($nums[$mid] > $target) {
352+
// 下次搜索在左区间:[$l,$mid-1]
353+
$r = $mid - 1;
354+
} else if ($nums[$mid] < $target) {
355+
// 下次搜索在右区间:[$mid+1,$r]
356+
$l = $mid + 1;
357+
} else {
358+
// 命中返回
359+
return $mid;
360+
}
361+
}
362+
return $r + 1;
363+
}
364+
```
321365

322366

323367
-----------------------

‎problems/0039.组合总和.md‎

Lines changed: 2 additions & 3 deletions
Original file line numberDiff line numberDiff line change
@@ -370,18 +370,17 @@ func backtracking(startIndex,sum,target int,candidates,trcak []int,res *[][]int)
370370
```js
371371
var combinationSum = function(candidates, target) {
372372
const res = [], path = [];
373-
candidates.sort(); // 排序
373+
candidates.sort((a,b)=>a-b); // 排序
374374
backtracking(0, 0);
375375
return res;
376376
function backtracking(j, sum) {
377-
if (sum > target) return;
378377
if (sum === target) {
379378
res.push(Array.from(path));
380379
return;
381380
}
382381
for(let i = j; i < candidates.length; i++ ) {
383382
const n = candidates[i];
384-
if(n > target - sum) continue;
383+
if(n > target - sum) break;
385384
path.push(n);
386385
sum += n;
387386
backtracking(i, sum);

‎problems/0040.组合总和II.md‎

Lines changed: 10 additions & 5 deletions
Original file line numberDiff line numberDiff line change
@@ -508,22 +508,27 @@ func backtracking(startIndex,sum,target int,candidates,trcak []int,res *[][]int)
508508
*/
509509
var combinationSum2 = function(candidates, target) {
510510
const res = []; path = [], len = candidates.length;
511-
candidates.sort();
511+
candidates.sort((a,b)=>a-b);
512512
backtracking(0, 0);
513513
return res;
514514
function backtracking(sum, i) {
515-
if (sum > target) return;
516515
if (sum === target) {
517516
res.push(Array.from(path));
518517
return;
519518
}
520-
let f = -1;
521519
for(let j = i; j < len; j++) {
522520
const n = candidates[j];
523-
if(n > target - sum || n === f) continue;
521+
if(j > i && candidates[j] === candidates[j-1]){
522+
//若当前元素和前一个元素相等
523+
//则本次循环结束,防止出现重复组合
524+
continue;
525+
}
526+
//如果当前元素值大于目标值-总和的值
527+
//由于数组已排序,那么该元素之后的元素必定不满足条件
528+
//直接终止当前层的递归
529+
if(n > target - sum) break;
524530
path.push(n);
525531
sum += n;
526-
f = n;
527532
backtracking(sum, j + 1);
528533
path.pop();
529534
sum -= n;

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