Google – Arbitrary Tree Longest Consecutive Path


Google – Arbitrary Tree Longest Consecutive Path
有两种题型:
1 只算从上到下,像leetcode298 Binary Tree Longest Consecutive Sequence
2 任意path
注意这里tree是任何的tree,不是binary tree.
[Solution]
第一题就是简单的dfs,简单到面试碰到原题还写出bug。他妈的…
第二题倒是非常复杂,对于每个node,要求4条最长路径:
1. 最长的递增path
2. 最长的递减path
3. 从左到右最长的path
4. 从右到左最长的path
递归return的是1,2两条,可以Wrap一个Pair class,也可以用HashMap.

  public int maxSumSubmatrix(int[][] matrix, int k) {
    if (matrix == null || matrix.length == 0) {
      return 0;
    }
    int m = matrix.length;
    int n = matrix[0].length;
    int result = Integer.MIN_VALUE;
    for (int i = 0; i < n; i++) {
      int r = i;
      int[] local = new int[m];
      while (r < n) {
        for (int x = 0; x < m; x++) {
          local[x] += matrix[x][r];
        }
        int localResult = maxSubArray(local, k);
        result = Math.max(result, localResult);
        r++;
      }
    }
    return result;
  }
  // maximum subarray less than k
  // can not use two pointer, [2, -7, 6, -1], k = 0
  // TreeSet, find ceiling, O(nlogn)
  private int maxSubArray(int[] nums, int k) {
    int result = Integer.MIN_VALUE;
    TreeSet<Integer> tSet = new TreeSet<>();
    int[] sums = new int[nums.length];
    sums[0] = nums[0];
    tSet.add(sums[0]);
    if (sums[0] <= k) {
      result = Math.max(result, sums[0]);
    }
    for (int i = 1; i < nums.length; i++) {
      sums[i] = sums[i - 1] + nums[i];
      if (sums[i] <= k) {
        result = Math.max(result, sums[i]);
      }
      Integer ceil = tSet.ceiling(sums[i] - k);
      if (ceil != null) {
        result = Math.max(result, sums[i] - ceil);
      }
      tSet.add(sums[i]);
    }
    return result;
  }
  // TreeSet find floor
  private int maxSubArray2(int[] nums, int k) {
    int n = nums.length;
    int result = Integer.MIN_VALUE;
    TreeSet<Integer> tSet = new TreeSet<>();
    int[] sums = new int[n];
    int sum = 0;
    for (int i = 0; i < n; i++) {
      sum += nums[i];
    }
    for (int i = n - 1; i >= 0; i--) {
      sums[i] = sum;
      if (sums[i] <= k) {
        result = Math.max(result, sums[i]);
      }
      if (i != n - 1) {
        Integer floor = tSet.floor(sums[i] + k);
        if (floor != null) {
          result = Math.max(result, floor - sums[i]);
        }
      }
      sum -= nums[i];
      tSet.add(sums[i]);
    }
    return result;
  }
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