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feat: add Java solutions to lc problems: No.2873~2876 (doocs#1736)
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‎solution/2800-2899/2873.Maximum Value of an Ordered Triplet I/README.md

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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 long maximumTripletValue(int[] nums) {
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long max, maxDiff, ans;
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max = 0;
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maxDiff = 0;
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ans = 0;
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for (int num : nums) {
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ans = Math.max(ans, num * maxDiff);
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max = Math.max(max, num);
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maxDiff = Math.max(maxDiff, max - num);
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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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‎solution/2800-2899/2873.Maximum Value of an Ordered Triplet I/README_EN.md

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@@ -58,6 +58,20 @@ It can be shown that there are no ordered triplets of indices with a value great
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### **Java**
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```java
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class Solution {
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public long maximumTripletValue(int[] nums) {
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long max, maxDiff, ans;
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max = 0;
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maxDiff = 0;
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ans = 0;
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for (int num : nums) {
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ans = Math.max(ans, num * maxDiff);
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max = Math.max(max, num);
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maxDiff = Math.max(maxDiff, max - num);
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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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‎solution/2800-2899/2873.Maximum Value of an Ordered Triplet I/Solution.java

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class Solution {
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public long maximumTripletValue(int[] nums) {
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long max, maxDiff, ans;
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max = 0;
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maxDiff = 0;
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ans = 0;
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for (int num : nums) {
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ans = Math.max(ans, num * maxDiff);
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max = Math.max(max, num);
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maxDiff = Math.max(maxDiff, max - num);
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}
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return ans;
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}
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}

‎solution/2800-2899/2874.Maximum Value of an Ordered Triplet II/README.md

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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 long maximumTripletValue(int[] nums) {
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long max, maxDiff, ans;
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max = 0;
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maxDiff = 0;
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ans = 0;
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for (int num : nums) {
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ans = Math.max(ans, num * maxDiff);
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max = Math.max(max, num);
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maxDiff = Math.max(maxDiff, max - num);
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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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‎solution/2800-2899/2874.Maximum Value of an Ordered Triplet II/README_EN.md

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@@ -58,6 +58,20 @@ It can be shown that there are no ordered triplets of indices with a value great
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### **Java**
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```java
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class Solution {
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public long maximumTripletValue(int[] nums) {
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long max, maxDiff, ans;
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max = 0;
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maxDiff = 0;
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ans = 0;
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for (int num : nums) {
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ans = Math.max(ans, num * maxDiff);
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max = Math.max(max, num);
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maxDiff = Math.max(maxDiff, max - num);
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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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‎solution/2800-2899/2874.Maximum Value of an Ordered Triplet II/Solution.java

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class Solution {
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public long maximumTripletValue(int[] nums) {
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long max, maxDiff, ans;
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max = 0;
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maxDiff = 0;
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ans = 0;
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for (int num : nums) {
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ans = Math.max(ans, num * maxDiff);
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max = Math.max(max, num);
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maxDiff = Math.max(maxDiff, max - num);
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}
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return ans;
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}
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}

‎solution/2800-2899/2875.Minimum Size Subarray in Infinite Array/README.md

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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 shortestSubarray(int[] nums, int k) {
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int n = nums.length;
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int minLength = n * 2 + 1;
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int l = 0;
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int sum = 0;
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for (int r = 0; r < n * 2; r++) {
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int start = l % n;
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int end = r % n;
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sum += nums[end];
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while (sum > k && l <= r) {
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start = l % n;
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sum -= nums[start];
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l++;
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}
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if (sum == k) {
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minLength = Math.min(minLength, r - l + 1);
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start = l % n;
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sum -= nums[start];
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l++;
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}
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}
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return minLength == n * 2 + 1 ? -1 : minLength;
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}
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public int minSizeSubarray(int[] nums, int target) {
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int n = nums.length;
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int sum = 0;
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for (int num : nums) {
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sum += num;
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}
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int k = target % sum;
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int ans = target / sum * n;
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if (k == 0) {
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return ans;
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}
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int res = shortestSubarray(nums, k);
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return res == -1 ? -1 : ans + res;
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}
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}
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```
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‎solution/2800-2899/2875.Minimum Size Subarray in Infinite Array/README_EN.md

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### **Java**
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```java
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class Solution {
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public int shortestSubarray(int[] nums, int k) {
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int n = nums.length;
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int minLength = n * 2 + 1;
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int l = 0;
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int sum = 0;
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for (int r = 0; r < n * 2; r++) {
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int start = l % n;
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int end = r % n;
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sum += nums[end];
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while (sum > k && l <= r) {
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start = l % n;
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sum -= nums[start];
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l++;
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}
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if (sum == k) {
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minLength = Math.min(minLength, r - l + 1);
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start = l % n;
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sum -= nums[start];
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l++;
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}
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}
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return minLength == n * 2 + 1 ? -1 : minLength;
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}
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public int minSizeSubarray(int[] nums, int target) {
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int n = nums.length;
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int sum = 0;
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for (int num : nums) {
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sum += num;
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}
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int k = target % sum;
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int ans = target / sum * n;
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if (k == 0) {
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return ans;
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}
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int res = shortestSubarray(nums, k);
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return res == -1 ? -1 : ans + res;
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}
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}
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```
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‎solution/2800-2899/2875.Minimum Size Subarray in Infinite Array/Solution.java

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class Solution {
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public int shortestSubarray(int[] nums, int k) {
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int n = nums.length;
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int minLength = n * 2 + 1;
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int l = 0;
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int sum = 0;
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for (int r = 0; r < n * 2; r++) {
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int start = l % n;
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int end = r % n;
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sum += nums[end];
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while (sum > k && l <= r) {
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start = l % n;
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sum -= nums[start];
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l++;
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}
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if (sum == k) {
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minLength = Math.min(minLength, r - l + 1);
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start = l % n;
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sum -= nums[start];
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l++;
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}
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}
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return minLength == n * 2 + 1 ? -1 : minLength;
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}
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public int minSizeSubarray(int[] nums, int target) {
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int n = nums.length;
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int sum = 0;
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for (int num : nums) {
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sum += num;
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}
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int k = target % sum;
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int ans = target / sum * n;
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if (k == 0) {
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return ans;
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}
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int res = shortestSubarray(nums, k);
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return res == -1 ? -1 : ans + res;
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}
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}

‎solution/2800-2899/2876.Count Visited Nodes in a Directed Graph/README.md

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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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void dfs(int curr, List<Integer> edges, int[] ans) {
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List<Integer> path = new ArrayList<>();
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int prev = -1;
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while (ans[curr] == 0) {
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path.add(curr);
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ans[curr] = prev == -1 ? -1 : ans[prev] - 1;
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prev = curr;
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curr = edges.get(curr);
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}
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int idx = path.size() - 1;
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if (ans[curr] < 0) {
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int cycle = ans[curr] - ans[path.get(idx)] + 1;
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int start = ans[curr];
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for (; idx >= 0 && ans[path.get(idx)] <= start; idx--) {
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ans[path.get(idx)] = cycle;
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}
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}
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for (; idx >= 0; idx--) {
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ans[path.get(idx)] = ans[edges.get(path.get(idx))] + 1;
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}
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}
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public int[] countVisitedNodes(List<Integer> edges) {
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int n = edges.size();
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int[] ans = new int[n];
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for (int i = 0; i < n; i++) {
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if (ans[i] > 0) {
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continue;
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}
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dfs(i, edges, ans);
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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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