feat: add python and java solutions to lcof question

yanglbme · yanglbme · commit c6c14e91092f · 2020-03-25T10:37:19.000+08:00
添加《剑指 Offer》题解：面试题33. 二叉搜索树的后序遍历序列
diff --git a/lcci/17.16.The Masseuse/README.md b/lcci/17.16.The Masseuse/README.md
@@ -32,7 +32,7 @@
 
 ## 解法
 <!-- 这里可写通用的实现逻辑 -->
-动态规划求解。
+递推求解。
 
 
 ### Python3
@@ -46,11 +46,12 @@ class Solution:
         n = len(nums)
         if n < 2:
             return nums[0]
-        dp = [0 for _ in range(n)]
-        dp[0], dp[1] = nums[0], max(nums[0], nums[1])
+        a, b = nums[0], max(nums[0], nums[1])
+        res = b
         for i in range(2, n):
-            dp[i] = max(dp[i - 1], dp[i - 2] + nums[i])
-        return dp[n - 1]
+            res = max(a + nums[i], b)
+            a, b = b, res
+        return res
 ```
 
 ### Java
@@ -66,13 +67,14 @@ class Solution {
         if (n < 2) {
             return n == 0 ? 0 : nums[0];
         }
-        int[] dp = new int[n];
-        dp[0] = nums[0];
-        dp[1] = Math.max(nums[0], nums[1]);
+        int a = nums[0], b = Math.max(nums[0], nums[1]);
+        int res = b;
         for (int i = 2; i < n; ++i) {
-            dp[i] = Math.max(dp[i - 2] + nums[i], dp[i - 1]);
+            res = Math.max(a + nums[i], b);
+            a = b;
+            b = res;
         }
-        return dp[n - 1];
+        return res;
     }
 }
 ```
diff --git a/lcci/17.16.The Masseuse/README_EN.md b/lcci/17.16.The Masseuse/README_EN.md
@@ -75,11 +75,12 @@ class Solution:
         n = len(nums)
         if n < 2:
             return nums[0]
-        dp = [0 for _ in range(n)]
-        dp[0], dp[1] = nums[0], max(nums[0], nums[1])
+        a, b = nums[0], max(nums[0], nums[1])
+        res = b
         for i in range(2, n):
-            dp[i] = max(dp[i - 1], dp[i - 2] + nums[i])
-        return dp[n - 1]
+            res = max(a + nums[i], b)
+            a, b = b, res
+        return res
 ```
 
 ### Java
@@ -94,13 +95,14 @@ class Solution {
         if (n < 2) {
             return n == 0 ? 0 : nums[0];
         }
-        int[] dp = new int[n];
-        dp[0] = nums[0];
-        dp[1] = Math.max(nums[0], nums[1]);
+        int a = nums[0], b = Math.max(nums[0], nums[1]);
+        int res = b;
         for (int i = 2; i < n; ++i) {
-            dp[i] = Math.max(dp[i - 2] + nums[i], dp[i - 1]);
+            res = Math.max(a + nums[i], b);
+            a = b;
+            b = res;
         }
-        return dp[n - 1];
+        return res;
     }
 }
 ```
diff --git a/lcci/17.16.The Masseuse/Solution.java b/lcci/17.16.The Masseuse/Solution.java
@@ -7,12 +7,13 @@ public int massage(int[] nums) {
         if (n < 2) {
             return n == 0 ? 0 : nums[0];
         }
-        int[] dp = new int[n];
-        dp[0] = nums[0];
-        dp[1] = Math.max(nums[0], nums[1]);
+        int a = nums[0], b = Math.max(nums[0], nums[1]);
+        int res = b;
         for (int i = 2; i < n; ++i) {
-            dp[i] = Math.max(dp[i - 2] + nums[i], dp[i - 1]);
+            res = Math.max(a + nums[i], b);
+            a = b;
+            b = res;
         }
-        return dp[n - 1];
+        return res;
     }
 }
diff --git a/lcci/17.16.The Masseuse/Solution.py b/lcci/17.16.The Masseuse/Solution.py
@@ -5,8 +5,9 @@ def massage(self, nums: List[int]) -> int:
         n = len(nums)
         if n < 2:
             return nums[0]
-        dp = [0 for _ in range(n)]
-        dp[0], dp[1] = nums[0], max(nums[0], nums[1])
+        a, b = nums[0], max(nums[0], nums[1])
+        res = b
         for i in range(2, n):
-            dp[i] = max(dp[i - 1], dp[i - 2] + nums[i])
-        return dp[n - 1]
+            res = max(a + nums[i], b)
+            a, b = b, res
+        return res
diff --git a/lcof/面试题33. 二叉搜索树的后序遍历序列/README.md b/lcof/面试题33. 二叉搜索树的后序遍历序列/README.md
@@ -0,0 +1,96 @@
+# [面试题33. 二叉搜索树的后序遍历序列](https://leetcode-cn.com/problems/er-cha-sou-suo-shu-de-hou-xu-bian-li-xu-lie-lcof/)
+
+## 题目描述
+<!-- 这里写题目描述 -->
+输入一个整数数组，判断该数组是不是某二叉搜索树的后序遍历结果。如果是则返回 `true`，否则返回 `false`。假设输入的数组的任意两个数字都互不相同。
+
+参考以下这颗二叉搜索树：
+
+```
+     5
+    / \
+   2   6
+  / \
+ 1   3
+```
+
+**示例 1：**
+
+```
+输入: [1,6,3,2,5]
+输出: false
+```
+
+**示例 2：**
+
+```
+输入: [1,3,2,6,5]
+输出: true
+```
+
+**提示：**
+
+- `数组长度 <= 1000`
+
+## 解法
+<!-- 这里可写通用的实现逻辑 -->
+二叉搜索树的后序遍历序列是 `[左子树, 右子树, 根结点]`，且左子树结点值均小于根结点，右子树结点值均大于根结点，递归判断即可。
+
+
+### Python3
+<!-- 这里可写当前语言的特殊实现逻辑 -->
+
+```python
+class Solution:
+    def verifyPostorder(self, postorder: List[int]) -> bool:
+        n = len(postorder)
+        if n < 2:
+            return True
+        for i in range(n):
+            if postorder[i] > postorder[-1]:
+                break
+        for j in range(i + 1, n - 1):
+            if postorder[j] < postorder[-1]:
+                return False
+        return (i == 0 or self.verifyPostorder(postorder[:i])) and (i == n - 1 or self.verifyPostorder(postorder[i:-1]))
+        
+```
+
+### Java
+<!-- 这里可写当前语言的特殊实现逻辑 -->
+
+```java
+class Solution {
+    public boolean verifyPostorder(int[] postorder) {
+        if (postorder.length == 0) {
+            return true;
+        }
+        return verifyPostorder(postorder, 0, postorder.length - 1);
+    }
+
+    private boolean verifyPostorder(int[] postorder, int from, int to) {
+        if (from == to) {
+            return true;
+        }
+        int i = from, j = from;
+        for (; i < to; ++i) {
+            if (postorder[i] > postorder[to]) {
+                break;
+            }
+        }
+        for (j = i + 1; j < to; ++j) {
+            if (postorder[j] < postorder[to]) {
+                return false;
+            }
+        }
+        return (i == from || verifyPostorder(postorder, from, i - 1)) && (i == to || verifyPostorder(postorder, i, to - 1));
+
+
+    }
+}
+```
+
+### ...
+```
+
+```
diff --git a/lcof/面试题33. 二叉搜索树的后序遍历序列/Solution.java b/lcof/面试题33. 二叉搜索树的后序遍历序列/Solution.java
@@ -0,0 +1,28 @@
+class Solution {
+    public boolean verifyPostorder(int[] postorder) {
+        if (postorder.length == 0) {
+            return true;
+        }
+        return verifyPostorder(postorder, 0, postorder.length - 1);
+    }
+
+    private boolean verifyPostorder(int[] postorder, int from, int to) {
+        if (from == to) {
+            return true;
+        }
+        int i = from, j = from;
+        for (; i < to; ++i) {
+            if (postorder[i] > postorder[to]) {
+                break;
+            }
+        }
+        for (j = i + 1; j < to; ++j) {
+            if (postorder[j] < postorder[to]) {
+                return false;
+            }
+        }
+        return (i == from || verifyPostorder(postorder, from, i - 1))
+                && (i == to || verifyPostorder(postorder, i, to - 1));
+
+    }
+}
diff --git a/lcof/面试题33. 二叉搜索树的后序遍历序列/Solution.py b/lcof/面试题33. 二叉搜索树的后序遍历序列/Solution.py
@@ -0,0 +1,12 @@
+class Solution:
+    def verifyPostorder(self, postorder: List[int]) -> bool:
+        n = len(postorder)
+        if n < 2:
+            return True
+        for i in range(n):
+            if postorder[i] > postorder[-1]:
+                break
+        for j in range(i + 1, n - 1):
+            if postorder[j] < postorder[-1]:
+                return False
+        return (i == 0 or self.verifyPostorder(postorder[:i])) and (i == n - 1 or self.verifyPostorder(postorder[i:-1]))

Original file line number	Diff line number	Diff line change
`@@ -7,12 +7,13 @@ public int massage(int[] nums) {`
`7`	`7`	`if (n < 2) {`
`8`	`8`	`return n == 0 ? 0 : nums[0];`
`9`	`9`	`}`
`10`		`- int[] dp = new int[n];`
`11`		`- dp[0] = nums[0];`
`12`		`- dp[1] = Math.max(nums[0], nums[1]);`
	`10`	`+ int a = nums[0], b = Math.max(nums[0], nums[1]);`
	`11`	`+ int res = b;`
`13`	`12`	`for (int i = 2; i < n; ++i) {`
`14`		`- dp[i] = Math.max(dp[i - 2] + nums[i], dp[i - 1]);`
	`13`	`+ res = Math.max(a + nums[i], b);`
	`14`	`+ a = b;`
	`15`	`+ b = res;`
`15`	`16`	`}`
`16`		`- return dp[n - 1];`
	`17`	`+ return res;`
`17`	`18`	`}`
`18`	`19`	`}`