Delete Node in the Middle of Singly Linked List

思路：
用当前节点next的值取代当前值。然后删掉node.next。一开始没想出来，这种题做一次就会了。

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776. Split BST

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Problem: Given a Binary Search Tree (BST) with root node  root , and a target value  V , split the tree into two subtrees where one subtree has nodes that are all smaller or equal to the target value, while the other subtree has all nodes that are greater than the target value.  It's not necessarily the case that the tree contains a node with value  V . Additionally, most of the structure of the original tree should remain.  Formally, for any child C with parent P in the original tree, if they are both in the same subtree after the split, then node C should still have the parent P. You should output the root TreeNode of both subtrees after splitting, in any order. Example 1: Input: root = [4,2,6,1,3,5,7], V = 2 Output: [[2,1],[4,3,6,null,null,5,7]] Explanation: Note that root, output[0], and output[1] are TreeNode objects, not arrays. The given tree [4,2,6,1,3,5,7] is represented by the following diagram: 4 / \ ...
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663. Equal Tree Partition

Problem: Given a binary tree with  n  nodes, your task is to check if it's possible to partition the tree to two trees which have the equal sum of values after removing  exactly  one edge on the original tree. Example 1: Input: 5 / \ 10 10 / \ 2 3 Output: True Explanation: 5 / 10 Sum: 15 10 / \ 2 3 Sum: 15 Example 2: Input: 1 / \ 2 10 / \ 2 20 Output: False Explanation: You can't split the tree into two trees with equal sum after removing exactly one edge on the tree. Note: The range of tree node value is in the range of [-100000, 100000]. 1 <= n <= 10000 Analysis: Get all sums, check whether the collections of sums contains totalSum/2. Do not include total sum to set of sums, if the total sum is 0, it will qualify.  Solution: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 class Solution { public boolean checkEqualTree(TreeNode ro...
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532. K-diff Pairs in an Array

Problem: Given an array of integers and an integer  k , you need to find the number of  unique  k-diff pairs in the array. Here a  k-diff  pair is defined as an integer pair (i, j), where  i  and  j  are both numbers in the array and their  absolute difference  is  k . Example 1: Input: [3, 1, 4, 1, 5], k = 2 Output: 2 Explanation: There are two 2-diff pairs in the array, (1, 3) and (3, 5). Although we have two 1s in the input, we should only return the number of unique pairs. Example 2: Input: [1, 2, 3, 4, 5], k = 1 Output: 4 Explanation: There are four 1-diff pairs in the array, (1, 2), (2, 3), (3, 4) and (4, 5). Example 3: Input: [1, 3, 1, 5, 4], k = 0 Output: 1 Explanation: There is one 0-diff pair in the array, (1, 1). Analysis: 通过这道题可以总结一条规律。同向双指针fast pointer不需要去重。 Solution: class Solution { public int findPairs ( int [] nums, int k) { if (nums == null || nums . length =...
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