上岸算法LeetCode Weekly Contest 270解题报告

【 NO.1 找出 3 位偶数】

解题思路
签到题，枚举所有组合即可。

代码展示

class Solution {
public int[] findEvenNumbers(int[] digits) {

   Set<Integer> result = new HashSet<>();
   for (int i = 0; i < digits.length; i++) {
       for (int j = 0; j < digits.length; j++) {
           for (int k = 0; k < digits.length; k++) {
               if (i == j || j == k || i == k) {
                   continue;
              }
               if (digits[i] != 0 && digits[k] % 2 == 0) {
                   result.add(digits[i] * 100 + digits[j] * 10 + digits[k]);
              }
          }
      }
  }
   int[] arr = result.stream().mapToInt(i -> i).toArray();
   Arrays.sort(arr);
   return arr;

}
}

【 NO.2 删除链表的中间节点】

解题思路
快慢指针的经典题目。

代码展示

class Solution {
public ListNode deleteMiddle(ListNode head) {

   if (head == null || head.next == null) {
       return null;
  }
   ListNode slow = head;
   ListNode fast = head.next;
   while (fast != null) {
       fast = fast.next;
       if (fast != null && fast.next != null) {
           slow = slow.next;
           fast = fast.next;
      }
  }
   slow.next = slow.next.next;
   return head;

}
}

【 NO.3 从二叉树一个节点到另一个节点每一步的方向】

解题思路
分别求出从根节点到 startValue 和 destValue 的路径，然后删去公共的部分，再把走向 startValue 的部分全部替换为 U 即可。

代码展示

class Solution {
public String getDirections(TreeNode root, int startValue, int destValue) {

   StringBuilder start = new StringBuilder();
   StringBuilder dest = new StringBuilder();
   getDirections(root, startValue, start);
   getDirections(root, destValue, dest);
   int common = 0;
   while (common < Math.min(start.length(), dest.length()) && start.charAt(common) == dest.charAt(common)) {
       common++;
  }
   if (common > 0) {
       start.delete(0, common);
       dest.delete(0, common);
  }
   for (int i = 0; i < start.length(); i++) {
       start.setCharAt(i, 'U');
  }
   return start.append(dest).toString();

}

private boolean getDirections(TreeNode root, int value, StringBuilder sb) {

   if (root == null) {
       return false;
  }
   if (root.val == value) {
       return true;
  }
   int len = sb.length();
   sb.append('L');
   if (getDirections(root.left, value, sb)) {
       return true;
  }
   sb.delete(len, sb.length());
   sb.append('R');
   return getDirections(root.right, value, sb);

}
}

【 NO.4 合法重新排列数对】

解题思路
有向图求欧拉路径的模板题。

代码展示

class Solution {
public int[][] validArrangement(int[][] pairs) {

   Map<Integer, LinkedList<Integer>> graph = new HashMap<>();
   Map<Integer, Integer> degree = new HashMap<>();
   for (var p : pairs) {
       if (!graph.containsKey(p[0])) {
           graph.put(p[0], new LinkedList<>());
      }
       graph.get(p[0]).add(p[1]);
       degree.put(p[0], degree.getOrDefault(p[0], 0) - 1);
       degree.put(p[1], degree.getOrDefault(p[1], 0) + 1);
  }
   List<int[]> result = new ArrayList<>();
   for (var e : degree.entrySet()) {
       if (e.getValue() < 0) {
           dfs(e.getKey(), result, graph);
      }
  }
   if (result.isEmpty()) {
       dfs(pairs[0][0], result, graph);
  }
   int[][] arr = new int[result.size()][];
   for (int i = 0; i < result.size(); i++) {
       arr[i] = result.get(result.size() - i - 1);
  }
   return arr;

}

private void dfs(int start, List<int[]> result, Map<Integer, LinkedList<Integer>> graph) {

   var next = graph.get(start);
   while (next != null && !next.isEmpty()) {
       int to = next.poll();
       dfs(to, result, graph);
       result.add(new int[]{start, to});
  }

}
}