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Commit a470c51

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feat: add solutions to lc problem: No.3028 (doocs#2315)
No.3028.Ant on the Boundary
1 parent cbbb894 commit a470c51

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7 files changed

+146
-8
lines changed

7 files changed

+146
-8
lines changed

‎solution/3000-3099/3028.Ant on the Boundary/README.md‎

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@@ -61,24 +61,71 @@
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## 解法
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### 方法一
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### 方法一:前缀和
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根据题目描述,我们只需要计算 $nums$ 的所有前缀和中有多少个 0ドル$ 即可。
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时间复杂度 $O(n),ドル其中 $n$ 为 $nums$ 的长度。空间复杂度 $O(1)$。
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<!-- tabs:start -->
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```python
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class Solution:
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def returnToBoundaryCount(self, nums: List[int]) -> int:
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return sum(s == 0 for s in accumulate(nums))
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```
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```java
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class Solution {
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public int returnToBoundaryCount(int[] nums) {
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int ans = 0, s = 0;
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for (int x : nums) {
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s += x;
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if (s == 0) {
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++ans;
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}
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}
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return ans;
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}
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}
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```
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```cpp
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class Solution {
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public:
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int returnToBoundaryCount(vector<int>& nums) {
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int ans = 0, s = 0;
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for (int x : nums) {
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s += x;
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ans += s == 0;
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}
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return ans;
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}
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};
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```
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```go
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func returnToBoundaryCount(nums []int) (ans int) {
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s := 0
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for _, x := range nums {
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s += x
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if s == 0 {
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ans++
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}
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}
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return
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}
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```
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```ts
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function returnToBoundaryCount(nums: number[]): number {
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let [ans, s] = [0, 0];
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for (const x of nums) {
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s += x;
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ans += s === 0 ? 1 : 0;
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}
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return ans;
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}
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```
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<!-- tabs:end -->

‎solution/3000-3099/3028.Ant on the Boundary/README_EN.md‎

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Original file line numberDiff line numberDiff line change
@@ -57,24 +57,71 @@ The ant never returned to the boundary, so the answer is 0.
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## Solutions
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### Solution 1
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### Solution 1: Prefix Sum
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Based on the problem description, we only need to calculate how many zeros are in all prefix sums of `nums`.
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The time complexity is $O(n),ドル where $n$ is the length of `nums`. The space complexity is $O(1)$.
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<!-- tabs:start -->
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```python
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class Solution:
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def returnToBoundaryCount(self, nums: List[int]) -> int:
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return sum(s == 0 for s in accumulate(nums))
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```
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```java
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class Solution {
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public int returnToBoundaryCount(int[] nums) {
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int ans = 0, s = 0;
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for (int x : nums) {
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s += x;
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if (s == 0) {
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++ans;
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}
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}
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return ans;
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}
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}
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```
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```cpp
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class Solution {
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public:
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int returnToBoundaryCount(vector<int>& nums) {
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int ans = 0, s = 0;
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for (int x : nums) {
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s += x;
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ans += s == 0;
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}
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return ans;
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}
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};
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```
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```go
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func returnToBoundaryCount(nums []int) (ans int) {
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s := 0
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for _, x := range nums {
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s += x
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if s == 0 {
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ans++
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}
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}
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return
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}
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```
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```ts
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function returnToBoundaryCount(nums: number[]): number {
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let [ans, s] = [0, 0];
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for (const x of nums) {
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s += x;
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ans += s === 0 ? 1 : 0;
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}
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return ans;
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}
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```
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<!-- tabs:end -->
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class Solution {
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public:
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int returnToBoundaryCount(vector<int>& nums) {
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int ans = 0, s = 0;
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for (int x : nums) {
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s += x;
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ans += s == 0;
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}
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return ans;
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}
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};
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Original file line numberDiff line numberDiff line change
@@ -0,0 +1,10 @@
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func returnToBoundaryCount(nums []int) (ans int) {
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s := 0
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for _, x := range nums {
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s += x
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if s == 0 {
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ans++
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}
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}
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return
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}
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@@ -0,0 +1,12 @@
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class Solution {
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public int returnToBoundaryCount(int[] nums) {
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int ans = 0, s = 0;
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for (int x : nums) {
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s += x;
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if (s == 0) {
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++ans;
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}
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}
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return ans;
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}
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}
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class Solution:
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def returnToBoundaryCount(self, nums: List[int]) -> int:
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return sum(s == 0 for s in accumulate(nums))
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function returnToBoundaryCount(nums: number[]): number {
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let [ans, s] = [0, 0];
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for (const x of nums) {
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s += x;
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ans += s === 0 ? 1 : 0;
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}
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return ans;
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}

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