add

5.leetcode/src/com/fanxb/common/Q198.java

@@ -0,0 +1,20 @@
+package com.fanxb.common;
+
+public class Q198 {
+    public int rob(int[] nums) {
+        //定义fn[k]表示有k个房子时的最大金额
+        int[] fn = new int[nums.length + 1];
+        //显然fn[0]=0,fn[1]=nums[0]
+        fn[1] = nums[0];
+        int res = fn[1];
+        for (int i = 2; i <= nums.length; i++) {
+            //当发现新房子时要获得最大收益有两种选择，偷这个新房子（不能偷前一个房子）或者不偷这一个房子（这样就能偷前一个房子了）,这两个情况用公式表示如下：
+            fn[i] = Math.max(fn[i - 1], fn[i - 2] + nums[i - 1]);
+            if (fn[i] > res) {
+                res = fn[i];
+            }
+        }
+        return res;
+    }
+
+}

5.leetcode/src/com/fanxb/common/Q213.java

@@ -14,13 +14,27 @@ import java.util.Arrays;
 public class Q213 {
 
     public int rob(int[] nums) {
-        int last = 0, cur = 0;
+        if (nums.length == 1) {
-        for (int num : nums) {
+            return nums[0];
-            int temp = cur;
+        } else {
-            cur = Math.max(cur, last + num);
+            return Math.max(robDo(Arrays.copyOfRange(nums, 0, nums.length - 1)), robDo(Arrays.copyOfRange(nums, 1, nums.length)));
-            last = temp;
         }
-        return cur;
+    }
+
+    private int robDo(int[] nums) {
+        //定义fn[k]表示有k个房子时的最大金额
+        int[] fn = new int[nums.length + 1];
+        //显然fn[0]=0,fn[1]=nums[0]
+        fn[1] = nums[0];
+        int res = fn[1];
+        for (int i = 2; i <= nums.length; i++) {
+            //当发现新房子时要获得最大收益有两种选择，偷这个新房子（不能偷前一个房子）或者不偷这一个房子（这样就能偷前一个房子了）,这两个情况用公式表示如下：
+            fn[i] = Math.max(fn[i - 1], fn[i - 2] + nums[i - 1]);
+            if (fn[i] > res) {
+                res = fn[i];
+            }
+        }
+        return res;
     }
 
     public static void main(String[] args) {

5.leetcode/src/com/fanxb/common/Q337.java

@@ -0,0 +1,46 @@
+package com.fanxb.common;
+
+import java.util.Arrays;
+import java.util.HashMap;
+import java.util.Map;
+
+/**
+ * 旋转链表
+ * 题目地址：https://leetcode-cn.com/problems/assign-cookies/submissions/
+ * 解题思路：
+ * 贪心算法，想不明白
+ *
+ * @author fanxb
+ * Date: 2020/6/9 15:10
+ */
+public class Q337 {
+
+
+    public int rob(TreeNode root) {
+        return robDo(root, new HashMap<>());
+    }
+
+    public int robDo(TreeNode root, Map<TreeNode, Integer> map) {
+        if (root == null) {
+            return 0;
+        }
+        Integer val = map.get(root);
+        if (val != null) {
+            return val;
+        }
+        //从根节点看起，根节点如果选择投了钱那么他的两个子节点都不能投钱，此是偷的钱的和为根节点的钱+从根节点孙子结点开始偷的钱的综合
+        //如果根节点不偷，那么他的两个子节点都可以投钱，此时偷钱的和为从两个子节点开始偷到的钱的综合
+        //然后再把子节点当作根节点来看就得到了递归关系
+        //sum1:两个孩子节点能偷到的钱
+        int sum1 = robDo(root.left, map) + robDo(root.right, map);
+        //四个孙子和roo节点能偷盗的钱
+        int sum2 = root.val + (root.left == null ? 0 : (robDo(root.left.left, map) + robDo(root.left.right, map))) + (root.right == null ? 0 : (robDo(root.right.left, map) + robDo(root.right.right, map)));
+        val = Math.max(sum1, sum2);
+        map.put(root, val);
+        return val;
+    }
+
+
+    public static void main(String[] args) {
+    }
+}

5.leetcode/src/com/fanxb/common/TreeNode.java

@@ -0,0 +1,20 @@
+package com.fanxb.common;
+
+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;
+    }
+}