動態規劃解決，比較當前位置目標和實際字符串的字母，再根據不同情況計算接下來的情形。

class Solution {public int longestCommonSubsequence(String text1, String text2) {char[] t1 = text1.toCharArray();char[] t2 = text2.toCharArray();int m = t1.length;int n = t2.length;int[][] dp = new int[m + 1][n + 1];for (int i = 0; i < m; i++) {for (int j = 0; j < n; j++) {dp[i + 1][j + 1] = t1[i] == t2[j] ? dp[i][j] + 1 : Math.max(dp[i][j + 1], dp[i + 1][j]);}}return dp[m][n];}
}

有可能刪除頭節點，所以要用到虛擬頭節點。
循環的過程中，將當前所指節點的下個節點的值取出作為標準，做到刪除過程不重不漏。

/*** Definition for singly-linked list.* public class ListNode {*     int val;*     ListNode next;*     ListNode() {}*     ListNode(int val) { this.val = val; }*     ListNode(int val, ListNode next) { this.val = val; this.next = next; }* }*/
class Solution {public ListNode deleteDuplicates(ListNode head) {ListNode dummy = new ListNode(0, head);ListNode cur = dummy;while (cur.next != null && cur.next.next != null) {int val = cur.next.val;if (cur.next.next.val == val) {while (cur.next != null && cur.next.val == val) {cur.next = cur.next.next;}} else {cur = cur.next;}}return dummy.next;}
}

這題困難的地方在于，需要將時間復雜度優化到 $O(log(m+n))$，必須用二分來解決。
目前先實現時間復雜度為 $O(m+n)$ 的方法，后續再完善。

class Solution {public double findMedianSortedArrays(int[] nums1, int[] nums2) {if (nums1.length > nums2.length) {int[] temp = nums1;nums1 = nums2;nums2 = temp;}int m = nums1.length;int n = nums2.length;int[] group1 = new int[m + 2];int[] group2 = new int[n + 2];group1[0] = group2[0] = Integer.MIN_VALUE;group1[m + 1] = group2[n + 1] = Integer.MAX_VALUE;System.arraycopy(nums1, 0, group1, 1, m);System.arraycopy(nums2, 0, group2, 1, n);int i = 0;int j = (m + n + 1) / 2;while (true) {if (group1[i] <= group2[j + 1] && group1[i + 1] > group2[j]) {int max = Math.max(group1[i], group2[j]);int min = Math.min(group1[i + 1], group2[j + 1]);return (m + n) % 2 == 1 ? max : (max + min) / 2.0;}i++;j--;}}
}

先序遍歷的變種，優先遍歷右子樹并維護深度，在結果中元素數量小于深度時，向其中添加新元素。

/*** Definition for a binary tree node.* public class TreeNode {*     int val;*     TreeNode left;*     TreeNode right;*     TreeNode() {}*     TreeNode(int val) { this.val = val; }*     TreeNode(int val, TreeNode left, TreeNode right) {*         this.val = val;*         this.left = left;*         this.right = right;*     }* }*/
class Solution {public List<Integer> rightSideView(TreeNode root) {List<Integer> res = new ArrayList<>();dfs(root, res, 0);return res;}private void dfs(TreeNode node, List<Integer> res, int depth) {if (node == null) {return;}if (res.size() == depth) {res.add(node.val);}dfs(node.right, res, depth + 1);dfs(node.left, res, depth + 1);}
}

模板題，理解的基礎上直接記。

/*** Definition for a binary tree node.* public class TreeNode {*     int val;*     TreeNode left;*     TreeNode right;*     TreeNode() {}*     TreeNode(int val) { this.val = val; }*     TreeNode(int val, TreeNode left, TreeNode right) {*         this.val = val;*         this.left = left;*         this.right = right;*     }* }*/
class Solution {public List<Integer> inorderTraversal(TreeNode root) {List<Integer> res = new ArrayList<>();dfs(root, res);return res;}private void dfs(TreeNode node, List<Integer> res) {if (node == null) {return;}dfs(node.left, res);res.add(node.val);dfs(node.right, res);}
}