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ShengyuanWangShengyuanWang
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feat: add solutions to lc problem: No.2210,2211 (#767)
- No.2210.Count Hills and Valleys in an Array - No.2211.Count Collisions on a Road
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‎solution/2200-2299/2210.Count Hills and Valleys in an Array/README.md

+42-2
Original file line numberDiff line numberDiff line change
@@ -56,22 +56,62 @@
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<!-- 这里可写通用的实现逻辑 -->
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先处理数组 `nums`，把连续的都变成一个，比如 `[6, 6, 5, 5, 4, 1]`，转换为 `[6, 5, 4, 1]` 之后，再进行比较。
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优化：
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可以使用双指针的方式，忽略相邻重复的元素，而无需改动原数组。
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def countHillValley(self, nums: List[int]) -> int:
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count = 0
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pt = 0
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while pt < len(nums)-1:
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if nums[pt] == nums[pt+1]:
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nums.pop(pt)
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continue
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else:
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pt += 1
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for i in range(1, len(nums)-1):
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if nums[i] < nums[i+1] and nums[i-1] > nums[i]:
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count += 1
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continue
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elif nums[i] > nums[i+1] and nums[i] > nums[i-1]:
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count += 1
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continue
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return count
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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public int countHillValley(int[] nums) {
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int count = 0;
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for(int i = 1,j = 0; i < nums.length-1; i++) {
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if(nums[i] == nums[i+1]) {
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continue;
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}
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if(nums[i] > nums[j] && nums[i] > nums[i+1]) {
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count++;
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}
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if(nums[i] < nums[j] && nums[i] < nums[i+1]){
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count++;
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}
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j = i;
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}
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return count;
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}
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}
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```
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### **TypeScript**

‎solution/2200-2299/2210.Count Hills and Valleys in an Array/README_EN.md

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Original file line numberDiff line numberDiff line change
@@ -54,15 +54,51 @@ There are 0 hills and valleys so we return 0.
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<!-- tabs:start -->
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### **Python3**
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First make the consecutive duplicate value to be unique with the side values.
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```python
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class Solution:
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def countHillValley(self, nums: List[int]) -> int:
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count = 0
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pt = 0
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while pt < len(nums)-1:
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if nums[pt] == nums[pt+1]:
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nums.pop(pt)
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continue
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else:
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pt += 1
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for i in range(1, len(nums)-1):
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if nums[i] < nums[i+1] and nums[i-1] > nums[i]:
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count += 1
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continue
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elif nums[i] > nums[i+1] and nums[i] > nums[i-1]:
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count += 1
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continue
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return count
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```
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### **Java**
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Use two pointers to solve the problem
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```java
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class Solution {
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public int countHillValley(int[] nums) {
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int count = 0;
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for(int i = 1,j = 0; i < nums.length-1; i++) {
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if(nums[i] == nums[i+1]) {
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continue;
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}
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if(nums[i] > nums[j] && nums[i] > nums[i+1]) {
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count++;
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}
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if(nums[i] < nums[j] && nums[i] < nums[i+1]){
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count++;
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}
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j = i;
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}
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return count;
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}
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}
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```
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### **TypeScript**

‎solution/2200-2299/2210.Count Hills and Valleys in an Array/Solution.java

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@@ -0,0 +1,18 @@
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class Solution {
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public int countHillValley(int[] nums) {
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int count = 0;
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for (int i = 1, j = 0; i < nums.length - 1; i++) {
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if(nums[i] == nums[i+1]) {
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continue;
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}
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if(nums[i] > nums[j] && nums[i] > nums[i+1]) {
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count++;
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}
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if(nums[i] < nums[j] && nums[i] < nums[i+1]){
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count++;
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}
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j = i;
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}
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return count;
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}
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}

‎solution/2200-2299/2210.Count Hills and Valleys in an Array/Solution.py

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Original file line numberDiff line numberDiff line change
@@ -0,0 +1,18 @@
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class Solution:
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def countHillValley(self, nums: List[int]) -> int:
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count = 0
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pt = 0
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while pt < len(nums)-1:
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if nums[pt] == nums[pt+1]:
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nums.pop(pt)
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continue
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else:
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pt += 1
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for i in range(1, len(nums)-1):
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if nums[i] < nums[i+1] and nums[i-1] > nums[i]:
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count += 1
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continue
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elif nums[i] > nums[i+1] and nums[i] > nums[i-1]:
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count += 1
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continue
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return count

‎solution/2200-2299/2211.Count Collisions on a Road/README.md

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Original file line numberDiff line numberDiff line change
@@ -57,6 +57,7 @@
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## 解法
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<!-- 这里可写通用的实现逻辑 -->
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这个题比较有意思，最后规律是左边向左的车辆可以出去，右边向右的车辆可以出去，中间不是S的都出不来
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<!-- tabs:start -->
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def countCollisions(self, directions: str) -> int:
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l, r = 0, len(directions)-1
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while l <= r and directions[l] == 'L':
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l += 1
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while l <= r and directions[r] == 'R':
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r -= 1
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count = 0
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for i in range(l, r+1):
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count += directions[i] != 'S'
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return count
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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public int countCollisions(String directions) {
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int l = 0, r = directions.length() -1, count = 0;
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while (l <= r && directions.substring(l, l+1).equals("L")) {
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l++;
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}
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while (l <= r && directions.substring(r, r+1).equals("R")) {
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r--;
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}
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for (int i = l; i <=r; i++) {
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if (!directions.substring(i, i+1).equals("S")) count += 1;
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}
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return count;
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}
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}
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```
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### **TypeScript**
@@ -82,10 +107,26 @@
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```
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85-
### **...**
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```
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### **C++**
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```cpp
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class Solution {
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public:
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int countCollisions(string directions) {
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int l = 0, r = directions.size() -1, count = 0;
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while (l <= r && directions[l] == 'L') {
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l++;
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}
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while (l <= r && directions[r] == 'R') {
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r--;
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}
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for (int i = l; i <=r; i++) {
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count += directions[i] != 'S';
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}
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return count;
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}
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};
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```
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<!-- tabs:end -->

‎solution/2200-2299/2211.Count Collisions on a Road/README_EN.md

+46-6
Original file line numberDiff line numberDiff line change
@@ -57,13 +57,37 @@ No cars will collide with each other. Thus, the total number of collisions that
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### **Python3**
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```python
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class Solution:
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def countCollisions(self, directions: str) -> int:
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l, r = 0, len(directions)-1
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while l <= r and directions[l] == 'L':
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l += 1
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while l <= r and directions[r] == 'R':
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r -= 1
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count = 0
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for i in range(l, r+1):
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count += directions[i] != 'S'
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return count
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```
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### **Java**
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```java
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class Solution {
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public int countCollisions(String directions) {
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int l = 0, r = directions.length() -1, count = 0;
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while (l <= r && directions.substring(l, l+1).equals("L")) {
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l++;
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}
82+
while (l <= r && directions.substring(r, r+1).equals("R")) {
83+
r--;
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}
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for (int i = l; i <=r; i++) {
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if (!directions.substring(i, i+1).equals("S")) count += 1;
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}
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return count;
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}
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}
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```
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### **TypeScript**
@@ -72,10 +96,26 @@ No cars will collide with each other. Thus, the total number of collisions that
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```
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### **...**
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```
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### **C++**
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```cpp
102+
class Solution {
103+
public:
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int countCollisions(string directions) {
105+
int l = 0, r = directions.size() -1, count = 0;
106+
while (l <= r && directions[l] == 'L') {
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l++;
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}
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while (l <= r && directions[r] == 'R') {
110+
r--;
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}
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for (int i = l; i <=r; i++) {
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count += directions[i] != 'S';
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}
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return count;
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}
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};
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```
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<!-- tabs:end -->

‎solution/2200-2299/2211.Count Collisions on a Road/Solution.cpp

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Original file line numberDiff line numberDiff line change
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class Solution {
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public:
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int countCollisions(string directions) {
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int l = 0, r = directions.size() -1, count = 0;
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while (l <= r && directions[l] == 'L') {
6+
l++;
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}
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while (l <= r && directions[r] == 'R') {
9+
r--;
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}
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for (int i = l; i <=r; i++) {
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count += directions[i] != 'S';
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}
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return count;
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}
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};

‎solution/2200-2299/2211.Count Collisions on a Road/Solution.java

+15
Original file line numberDiff line numberDiff line change
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1+
class Solution {
2+
public int countCollisions(String directions) {
3+
int l = 0, r = directions.length() -1, count = 0;
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while (l <= r && directions.substring(l, l+1).equals("L")) {
5+
l++;
6+
}
7+
while (l <= r && directions.substring(r, r+1).equals("R")) {
8+
r--;
9+
}
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for (int i = l; i <=r; i++) {
11+
if (!directions.substring(i, i+1).equals("S")) count += 1;
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}
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return count;
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}
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}

‎solution/2200-2299/2211.Count Collisions on a Road/Solution.py

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Original file line numberDiff line numberDiff line change
@@ -0,0 +1,11 @@
1+
class Solution:
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def countCollisions(self, directions: str) -> int:
3+
l, r = 0, len(directions)-1
4+
while l <= r and directions[l] == 'L':
5+
l += 1
6+
while l <= r and directions[r] == 'R':
7+
r -= 1
8+
count = 0
9+
for i in range(l, r+1):
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count += directions[i] != 'S'
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return count

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