From eb04b27e4119162316051cb1891ed7c58a838988 Mon Sep 17 00:00:00 2001
From: liu13 <1099976891@qq.com>
Date: Mon, 15 Apr 2019 10:31:43 +0800
Subject: [PATCH] 20190415

---
 code/lc1025.java | 15 +++++++++++++++
 code/lc1026.java | 47 +++++++++++++++++++++++++++++++++++++++++++++++
 code/lc1027.java | 24 ++++++++++++++++++++++++
 3 files changed, 86 insertions(+)
 create mode 100644 code/lc1025.java
 create mode 100644 code/lc1026.java
 create mode 100644 code/lc1027.java

diff --git a/code/lc1025.java b/code/lc1025.java
new file mode 100644
index 0000000..56a5a05
--- /dev/null
+++ b/code/lc1025.java
@@ -0,0 +1,15 @@
+package code;
+/*
+ * 1025. Divisor Game
+ * 题意：
+ * 难度：
+ * 分类：
+ * 思路：先选的有主动权
+ * Tips：
+ */
+public class lc1025 {
+    public boolean divisorGame(int N) {
+        if(N%2==0) return true;
+        return false;
+    }
+}
diff --git a/code/lc1026.java b/code/lc1026.java
new file mode 100644
index 0000000..6beb0cc
--- /dev/null
+++ b/code/lc1026.java
@@ -0,0 +1,47 @@
+package code;
+/*
+ * 1025. Divisor Game
+ * 题意：父节点减子节点的绝对值最大
+ * 难度：
+ * 分类：
+ * 思路：自己写的自下向上，返回的时候再计算。
+ *      可以自顶向下的，到叶子节点计算就可以
+ * Tips：
+ */
+public class lc1026 {
+    public class TreeNode {
+        int val;
+        TreeNode left;
+        TreeNode right;
+        TreeNode(int x) {
+            val = x;
+        }
+    }
+    int res_val = 0;
+    public int maxAncestorDiff(TreeNode root) {
+        helper(root);
+        return res_val;
+    }
+
+    public int[] helper(TreeNode root){
+        int[] res = new int[2]; //存一个最大，一个最小
+        if(root.left==null&&root.right==null){
+            res[0] = root.val;
+            res[1] = root.val;
+            return res;
+        }
+        int[] left = new int[2];
+        int[] right = new int[2];
+        left[0] = Integer.MIN_VALUE;
+        right[0] = Integer.MIN_VALUE;
+        left[1] = Integer.MAX_VALUE;
+        right[1] = Integer.MAX_VALUE;
+        if(root.left!=null) left = helper(root.left);   //可能为空，只有一边有节点
+        if(root.right!=null) right = helper(root.right);
+        res_val = Math.max(Math.abs(root.val - Math.max(left[0], right[0])), res_val);
+        res_val = Math.max(Math.abs(root.val - Math.min(left[1], right[1])), res_val);
+        res[0] = Math.max(Math.max(left[0], right[0]), root.val);   //别忘了和root节点本身的值比
+        res[1] = Math.min(Math.min(left[1], right[1]), root.val);
+        return res;
+    }
+}
diff --git a/code/lc1027.java b/code/lc1027.java
new file mode 100644
index 0000000..5404934
--- /dev/null
+++ b/code/lc1027.java
@@ -0,0 +1,24 @@
+package code;
+/*
+ * 1027. Longest Arithmetic Sequence
+ * 题意：最长等差数列
+ * 难度：Medium
+ * 分类：Dynamic Programming
+ * 思路：还是要在每个位置上都记录一下，不是说一个大的数组就可以了。因为相同长度的等差，后续数字往上接的时候不一定接哪一个，都要保存，并不是说一定去接小的那个。
+ *      二维dp, 一维记录位置，一维记录差分
+ * Tips：很棒的题
+ */
+public class lc1027 {
+    public static int longestArithSeqLength(int[] A) {
+        int res = 0;
+        int[][] dp = new int[A.length][20000];
+        for (int i = 0; i < A.length ; i++) {
+            for (int j = 0; j < i ; j++) {
+                int ind = A[i]-A[j]+10000;
+                dp[i][ind] = dp[j][ind] + 1;
+                res = Math.max(res, dp[i][ind]);
+            }
+        }
+        return res+1; //加1
+    }
+}
-- 
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