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Merge branch 'master' of github.com:youngyangyang04/leetcode-master

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programmercarl
2022-09-22 11:40:03 +08:00
parent 5a1841aa2b d11cf7b7ee
commit 721bab7636
6 changed files with 140 additions and 65 deletions
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20
problems/0035.搜索插入位置.md
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@ -270,6 +270,26 @@ func searchInsert(nums []int, target int) int {
}
```
### Rust
```rust
impl Solution {
pub fn search_insert(nums: Vec<i32>, target: i32) -> i32 {
let mut left = 0;
let mut right = nums.len();
while left < right {
let mid = (left + right) / 2;
match nums[mid].cmp(&target) {
Ordering::Less => left = mid + 1,
Ordering::Equal => return ((left + right) / 2) as i32,
Ordering::Greater => right = mid,
}
}
((left + right) / 2) as i32
}
}
```
### Python
```python
class Solution:

1
problems/0077.组合.md
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@ -368,7 +368,6 @@ class Solution {
### Python
Python2:
```python
class Solution(object):
def combine(self, n, k):

28
problems/0139.单词拆分.md
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@ -234,11 +234,13 @@ Java：
```java
class Solution {
public boolean wordBreak(String s, List<String> wordDict) {
HashSet<String> set = new HashSet<>(wordDict);
boolean[] valid = new boolean[s.length() + 1];
valid[0] = true;
for (int i = 1; i <= s.length(); i++) {
for (int j = 0; j < i; j++) {
if (wordDict.contains(s.substring(j,i)) && valid[j]) {
for (int j = 0; j < i && !valid[i]; j++) {
if (set.contains(s.substring(j, i)) && valid[j]) {
valid[i] = true;
}
}
@ -248,6 +250,26 @@ class Solution {
}
}
// 另一种思路的背包算法
class Solution {
public boolean wordBreak(String s, List<String> wordDict) {
boolean[] dp = new boolean[s.length() + 1];
dp[0] = true;
for (int i = 1; i <= s.length(); i++) {
for (String word : wordDict) {
int len = word.length();
if (i >= len && dp[i - len] && word.equals(s.substring(i - len, i))) {
dp[i] = true;
break;
}
}
}
return dp[s.length()];
}
}
// 回溯法+记忆化
class Solution {
private Set<String> set;
@ -285,7 +307,7 @@ class Solution {
Python：
```python3
```python
class Solution:
def wordBreak(self, s: str, wordDict: List[str]) -> bool:
'''排列'''

27
problems/0704.二分查找.md
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@ -220,7 +220,8 @@ class Solution:
（版本二）左闭右开区间
```class Solution:
```python
class Solution:
def search(self, nums: List[int], target: int) -> int:
if nums is None or len(nums)==0:
return -1
@ -516,6 +517,7 @@ impl Solution {
**C:**
```c
// (版本一) 左闭右闭区间 [left, right]
int search(int* nums, int numsSize, int target){
int left = 0;
int right = numsSize-1;
@ -541,6 +543,29 @@ int search(int* nums, int numsSize, int target){
return -1;
}
```
```C
// (版本二) 左闭右开区间 [left, right)
int search(int* nums, int numsSize, int target){
int length = numsSize;
int left = 0;
int right = length; //定义target在左闭右开的区间里，即：[left, right)
int middle = 0;
while(left < right){ // left == right时，区间[left, right)属于空集，所以用 < 避免该情况
int middle = left + (right - left) / 2;
if(nums[middle] < target){
//target位于(middle , right) 中为保证集合区间的左闭右开性，可等价为[middle + 1,right)
left = middle + 1;
}else if(nums[middle] > target){
//target位于[left, middle)中
right = middle ;
}else{ // nums[middle] == target ，找到目标值target
return middle;
}
}
//未找到目标值，返回-1
return -1;
}
```
**PHP:**
```php

127
problems/0707.设计链表.md
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@ -358,7 +358,7 @@ class MyLinkedList {
return;
}
ListNode pred = head;
for (int i = 0; i < index - 1; i++) {
for (int i = 0; i < index ; i++) {
pred = pred.next;
}
pred.next = pred.next.next;
@ -366,92 +366,101 @@ class MyLinkedList {
}
//双链表
class MyLinkedList {
class ListNode {
int val;
ListNode next,prev;
ListNode(int x) {val = x;}
class ListNode{
int val;
ListNode next,prev;
ListNode() {};
ListNode(int val){
this.val = val;
}
}
class MyLinkedList {
//记录链表中元素的数量
int size;
ListNode head,tail;//Sentinel node
/** Initialize your data structure here. */
//记录链表的虚拟头结点和尾结点
ListNode head,tail;
public MyLinkedList() {
size = 0;
head = new ListNode(0);
tail = new ListNode(0);
head.next = tail;
tail.prev = head;
//初始化操作
this.size = 0;
this.head = new ListNode(0);
this.tail = new ListNode(0);
//这一步非常关键，否则在加入头结点的操作中会出现null.next的错误！！！
head.next=tail;
tail.prev=head;
}
/** Get the value of the index-th node in the linked list. If the index is invalid, return -1. */
public int get(int index) {
if(index < 0 || index >= size){return -1;}
ListNode cur = head;
// 通过判断 index < (size - 1) / 2 来决定是从头结点还是尾节点遍历，提高效率
if(index < (size - 1) / 2){
for(int i = 0; i <= index; i++){
cur = cur.next;
}
}else{
//判断index是否有效
if(index<0 || index>=size){
return -1;
}
ListNode cur = this.head;
//判断是哪一边遍历时间更短
if(index >= size / 2){
//tail开始
cur = tail;
for(int i = 0; i <= size - index - 1; i++){
for(int i=0; i< size-index; i++){
cur = cur.prev;
}
}else{
for(int i=0; i<= index; i++){
cur = cur.next;
}
}
return cur.val;
}
/** Add a node of value val before the first element of the linked list. After the insertion, the new node will be the first node of the linked list. */
public void addAtHead(int val) {
ListNode cur = head;
ListNode newNode = new ListNode(val);
newNode.next = cur.next;
cur.next.prev = newNode;
cur.next = newNode;
newNode.prev = cur;
size++;
//等价于在第0个元素前添加
addAtIndex(0,val);
}
/** Append a node of value val to the last element of the linked list. */
public void addAtTail(int val) {
ListNode cur = tail;
ListNode newNode = new ListNode(val);
newNode.next = tail;
newNode.prev = cur.prev;
cur.prev.next = newNode;
cur.prev = newNode;
size++;
//等价于在最后一个元素(null)前添加
addAtIndex(size,val);
}
/** Add a node of value val before the index-th node in the linked list. If index equals to the length of linked list, the node will be appended to the end of linked list. If index is greater than the length, the node will not be inserted. */
public void addAtIndex(int index, int val) {
if(index > size){return;}
if(index < 0){index = 0;}
ListNode cur = head;
for(int i = 0; i < index; i++){
cur = cur.next;
//index大于链表长度
if(index>size){
return;
}
//index小于0
if(index<0){
index = 0;
}
ListNode newNode = new ListNode(val);
newNode.next = cur.next;
cur.next.prev = newNode;
newNode.prev = cur;
cur.next = newNode;
size++;
//找到前驱
ListNode pre = this.head;
for(int i=0; i<index; i++){
pre = pre.next;
}
//新建结点
ListNode newNode = new ListNode(val);
newNode.next = pre.next;
pre.next.prev = newNode;
newNode.prev = pre;
pre.next = newNode;
}
/** Delete the index-th node in the linked list, if the index is valid. */
public void deleteAtIndex(int index) {
if(index >= size || index < 0){return;}
ListNode cur = head;
for(int i = 0; i < index; i++){
cur = cur.next;
//判断索引是否有效
if(index<0 || index>=size){
return;
}
cur.next.next.prev = cur;
cur.next = cur.next.next;
//删除操作
size--;
ListNode pre = this.head;
for(int i=0; i<index; i++){
pre = pre.next;
}
pre.next.next.prev = pre;
pre.next = pre.next.next;
}
}

2
problems/0797.所有可能的路径.md
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@ -296,7 +296,7 @@ public:
### Java
### Python
```
```python
class Solution:
def __init__(self):
self.result = []