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feat: add solutions to lcof2 problems: No.008,009
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8 files changed

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‎lcof2/剑指 Offer II 008. 和大于等于 target 的最短子数组/README.md

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@@ -66,15 +66,74 @@
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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 minSubArrayLen(self, target: int, nums: List[int]) -> int:
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n = len(nums)
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ans = float('inf')
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sum = 0
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left, right = 0, 0
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while right < n:
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sum += nums[right]
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right += 1
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while sum >= target:
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ans = min(ans, right - left)
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sum -= nums[left]
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left += 1
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return 0 if ans == float('inf') else ans
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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 minSubArrayLen(int target, int[] nums) {
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int n = nums.length;
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int ans = Integer.MAX_VALUE;
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int sum = 0;
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int left = 0, right = 0;
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while (right < n) {
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sum += nums[right++];
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while (sum >= target) {
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ans = Math.min(ans, right - left);
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sum -= nums[left++];
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}
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}
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return ans == Integer.MAX_VALUE ? 0 : ans;
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}
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}
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```
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### **Go**
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```go
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func minSubArrayLen(target int, nums []int) int {
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n := len(nums)
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ans := math.MaxInt32
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sum := 0
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left, right := 0, 0
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for right < n {
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sum += nums[right]
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right++
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for sum >= target {
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ans = min(ans, right-left)
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sum -= nums[left]
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left++
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}
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}
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if ans == math.MaxInt32 {
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return 0
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}
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return ans
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}
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func min(a, b int) int {
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if a < b {
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return a
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}
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return b
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}
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```
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### **...**

‎lcof2/剑指 Offer II 008. 和大于等于 target 的最短子数组/Solution.go

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func minSubArrayLen(target int, nums []int) int {
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n := len(nums)
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ans := math.MaxInt32
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sum := 0
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left, right := 0, 0
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for right < n {
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sum += nums[right]
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right++
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for sum >= target {
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ans = min(ans, right-left)
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sum -= nums[left]
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left++
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}
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}
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if ans == math.MaxInt32 {
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return 0
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}
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return ans
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}
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func min(a, b int) int {
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if a < b {
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return a
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}
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return b
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}

‎lcof2/剑指 Offer II 008. 和大于等于 target 的最短子数组/Solution.java

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class Solution {
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public int minSubArrayLen(int target, int[] nums) {
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int n = nums.length;
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int ans = Integer.MAX_VALUE;
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int sum = 0;
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int left = 0, right = 0;
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while (right < n) {
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sum += nums[right++];
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while (sum >= target) {
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ans = Math.min(ans, right - left);
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sum -= nums[left++];
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}
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}
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return ans == Integer.MAX_VALUE ? 0 : ans;
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}
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}

‎lcof2/剑指 Offer II 008. 和大于等于 target 的最短子数组/Solution.py

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class Solution:
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def minSubArrayLen(self, target: int, nums: List[int]) -> int:
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n = len(nums)
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ans = float('inf')
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sum = 0
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left, right = 0, 0
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while right < n:
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sum += nums[right]
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right += 1
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while sum >= target:
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ans = min(ans, right - left)
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sum -= nums[left]
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left += 1
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return 0 if ans == float('inf') else ans

‎lcof2/剑指 Offer II 009. 乘积小于 K 的子数组/README.md

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@@ -42,22 +42,73 @@
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<!-- 这里可写通用的实现逻辑 -->
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利用滑动窗口，我们能求出每个不同 `right` 结尾的合法子数组的个数
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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 numSubarrayProductLessThanK(self, nums: List[int], k: int) -> int:
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n = len(nums)
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ans = 0
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sum = 1
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left, right = 0, 0
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while right < n:
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sum *= nums[right]
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right += 1
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while sum >= k and left < right:
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sum /= nums[left]
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left += 1
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ans += right - left
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return ans
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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 numSubarrayProductLessThanK(int[] nums, int k) {
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int n = nums.length;
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int ans = 0;
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int sum = 1;
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int left = 0, right = 0;
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while (right < n) {
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sum *= nums[right++];
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while (sum >= k && left < right) {
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sum /= nums[left++];
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}
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ans += right - left;
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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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```go
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func numSubarrayProductLessThanK(nums []int, k int) int {
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n := len(nums)
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ans := 0
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sum := 1
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left, right := 0, 0
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for right < n {
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sum *= nums[right]
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right++
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for sum >= k && left < right {
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sum /= nums[left]
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left++
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}
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ans += right - left
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}
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return ans
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}
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```
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### **...**

‎lcof2/剑指 Offer II 009. 乘积小于 K 的子数组/Solution.go

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func numSubarrayProductLessThanK(nums []int, k int) int {
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n := len(nums)
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ans := 0
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sum := 1
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left, right := 0, 0
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for right < n {
7+
sum *= nums[right]
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right++
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for sum >= k && left < right {
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sum /= nums[left]
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left++
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}
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ans += right - left
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}
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return ans
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}

‎lcof2/剑指 Offer II 009. 乘积小于 K 的子数组/Solution.java

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@@ -0,0 +1,16 @@
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class Solution {
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public int numSubarrayProductLessThanK(int[] nums, int k) {
3+
int n = nums.length;
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int ans = 0;
5+
int sum = 1;
6+
int left = 0, right = 0;
7+
while (right < n) {
8+
sum *= nums[right++];
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while (sum >= k && left < right) {
10+
sum /= nums[left++];
11+
}
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ans += right - left;
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}
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return ans;
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}
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}

‎lcof2/剑指 Offer II 009. 乘积小于 K 的子数组/Solution.py

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@@ -0,0 +1,14 @@
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class Solution:
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def numSubarrayProductLessThanK(self, nums: List[int], k: int) -> int:
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n = len(nums)
4+
ans = 0
5+
sum = 1
6+
left, right = 0, 0
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while right < n:
8+
sum *= nums[right]
9+
right += 1
10+
while sum >= k and left < right:
11+
sum /= nums[left]
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left += 1
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ans += right - left
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return ans

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