) -> bool {
+ let mut five = 0;
+ let mut ten = 0;
+ // let mut twenty = 0;
+ for bill in bills {
+ if bill == 5 { five += 1; }
+ if bill == 10 {
+ if five <= 0 { return false; }
+ ten += 1;
+ five -= 1;
+ }
+ if bill == 20 {
+ if five > 0 && ten > 0 {
+ five -= 1;
+ ten -= 1;
+ // twenty += 1;
+ } else if five >= 3 {
+ five -= 3;
+ // twenty += 1;
+ } else { return false; }
+ }
+ }
+ true
+ }
+}
+```
+
### C
```c
bool lemonadeChange(int* bills, int billsSize){
diff --git a/problems/0922.按奇偶排序数组II.md b/problems/0922.按奇偶排序数组II.md
index 49547a15..4e281a3c 100644
--- a/problems/0922.按奇偶排序数组II.md
+++ b/problems/0922.按奇偶排序数组II.md
@@ -147,6 +147,59 @@ class Solution {
}
```
+### java
+
+```java
+//方法一：采用额外的数组空间
+class Solution {
+ public int[] sortArrayByParityII(int[] nums) {
+ //定义结果数组 result
+ int[] result = new int[nums.length];
+ int even = 0, odd = 1;
+ for(int i = 0; i < nums.length; i++){
+ //如果为偶数
+ if(nums[i] % 2 == 0){
+ result[even] = nums[i];
+ even += 2;
+ }else{
+ result[odd] = nums[i];
+ odd += 2;
+ }
+ }
+ return result;
+ }
+}
+```
+```java
+//方法二：不采用额外的数组空间
+class Solution922 {
+ public int[] sortArrayByParityII(int[] nums) {
+ //定义双指针
+ int oddPoint = 1, evenPoint = 0;
+ //开始移动并交换，最后一层必然为相互交换后再移动或者相同直接移动
+ while(oddPoint < nums.length && evenPoint < nums.length){
+ //进行判断
+ if(nums[oddPoint] % 2 == 0 && nums[evenPoint] % 2 == 1){ //如果均不满足
+ int temp = 0;
+ temp = nums[oddPoint];
+ nums[oddPoint] = nums[evenPoint];
+ nums[evenPoint] = temp;
+ oddPoint += 2;
+ evenPoint += 2;
+ }else if(nums[oddPoint] % 2 == 0 && nums[evenPoint] % 2 == 0){ //偶数满足
+ evenPoint += 2;
+ }else if(nums[oddPoint] % 2 == 1 && nums[evenPoint] % 2 == 1){ //奇数满足
+ oddPoint += 2;
+ }else{
+ oddPoint += 2;
+ evenPoint += 2;
+ }
+ }
+ return nums;
+ }
+}
+```
+
### Python3
```python
diff --git a/problems/链表理论基础.md b/problems/链表理论基础.md
index 1a29c32a..8378f7f2 100644
--- a/problems/链表理论基础.md
+++ b/problems/链表理论基础.md
@@ -186,7 +186,7 @@ TypeScript:
```typescript
class ListNode {
public val: number;
- public next: ListNode = null;
+ public next: ListNode|null = null;
constructor(value: number) {
this.val = value;
this.next = null;
diff --git a/problems/面试题02.07.链表相交.md b/problems/面试题02.07.链表相交.md
index abf7a2f8..6d6810a3 100644
--- a/problems/面试题02.07.链表相交.md
+++ b/problems/面试题02.07.链表相交.md
@@ -5,7 +5,9 @@
参与本项目，贡献其他语言版本的代码，拥抱开源，让更多学习算法的小伙伴们收益！
-# 面试题 02.07. 链表相交
+# 面试题 02.07. 链表相交
+
+同：160.链表相交
[力扣题目链接](https://leetcode.cn/problems/intersection-of-two-linked-lists-lcci/)
diff --git a/添加0222.完全二叉树的节点个数Go版本.md b/添加0222.完全二叉树的节点个数Go版本.md
new file mode 100644
index 00000000..6001e7b7
--- /dev/null
+++ b/添加0222.完全二叉树的节点个数Go版本.md
@@ -0,0 +1,25 @@
+```go
+func countNodes(root *TreeNode) int {
+ if root == nil {
+ return 0
+ }
+ q := list.New()
+ q.PushBack(root)
+ res := 0
+ for q.Len() > 0 {
+ n := q.Len()
+ for i := 0; i < n; i++ {
+ node := q.Remove(q.Front()).(*TreeNode)
+ if node.Left != nil {
+ q.PushBack(node.Left)
+ }
+ if node.Right != nil {
+ q.PushBack(node.Right)
+ }
+ res++
+ }
+ }
+ return res
+}
+```
+
diff --git a/添加559.n叉树的最大深度Go版本.md b/添加559.n叉树的最大深度Go版本.md
new file mode 100644
index 00000000..3172837d
--- /dev/null
+++ b/添加559.n叉树的最大深度Go版本.md
@@ -0,0 +1,22 @@
+```go
+func maxDepth(root *Node) int {
+ if root == nil {
+ return 0
+ }
+ q := list.New()
+ q.PushBack(root)
+ depth := 0
+ for q.Len() > 0 {
+ n := q.Len()
+ for i := 0; i < n; i++ {
+ node := q.Remove(q.Front()).(*Node)
+ for j := range node.Children {
+ q.PushBack(node.Children[j])
+ }
+ }
+ depth++
+ }
+ return depth
+}
+```
+
