feat: add python and java solutions to leetcode problem: No.0198

yanglbme · yanglbme · commit ac3d1b7c4108 · 2020-12-14T16:52:40.000+08:00
diff --git a/solution/0100-0199/0127.Word Ladder/README_EN.md b/solution/0100-0199/0127.Word Ladder/README_EN.md
@@ -7,27 +7,18 @@
 <p>Given two words (<em>beginWord</em> and <em>endWord</em>), and a dictionary&#39;s word list, find the length of shortest transformation sequence from <em>beginWord</em> to <em>endWord</em>, such that:</p>
 
 <ol>
-
     <li>Only one letter can be changed at a time.</li>
-
     <li>Each transformed word must exist in the word list. Note that <em>beginWord</em> is <em>not</em> a transformed word.</li>
-
 </ol>
 
 <p><strong>Note:</strong></p>
 
 <ul>
-
     <li>Return 0 if there is no such transformation sequence.</li>
-
     <li>All words have the same length.</li>
-
     <li>All words contain only lowercase alphabetic characters.</li>
-
     <li>You may assume no duplicates in the word list.</li>
-
     <li>You may assume <em>beginWord</em> and <em>endWord</em> are non-empty and are not the same.</li>
-
 </ul>
 
 <p><strong>Example 1:</strong></p>
diff --git a/solution/0100-0199/0139.Word Break/README_EN.md b/solution/0100-0199/0139.Word Break/README_EN.md
@@ -9,11 +9,8 @@
 <p><strong>Note:</strong></p>
 
 <ul>
-
     <li>The same word in the dictionary may be reused multiple times in the segmentation.</li>
-
     <li>You may assume the dictionary does not contain duplicate words.</li>
-
 </ul>
 
 <p><strong>Example 1:</strong></p>
diff --git a/solution/0100-0199/0198.House Robber/README.md b/solution/0100-0199/0198.House Robber/README.md
@@ -35,15 +35,44 @@
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```python
-
+class Solution:
+    def rob(self, nums: List[int]) -> int:
+        if not nums:
+            return 0
+        n = len(nums)
+        if n == 1:
+            return nums[0]
+        dp = [0 for _ in range(n)]
+        dp[0] = nums[0]
+        dp[1] = max(nums[0], nums[1])
+        for i in range(2, n):
+            dp[i] = max(dp[i - 1], dp[i - 2] + nums[i])
+        return dp[n - 1]
 ```
 
 ### **Java**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```java
-
+class Solution {
+    public int rob(int[] nums) {
+        int n;
+        if (nums == null || (n = nums.length) == 0) {
+            return 0;
+        }
+        if (n == 1) {
+            return nums[0];
+        }
+        int[] dp = new int[n];
+        dp[0] = nums[0];
+        dp[1] = Math.max(nums[0], nums[1]);
+        for (int i = 2; i < n; ++i) {
+            dp[i] = Math.max(dp[i - 1], dp[i - 2] + nums[i]);
+        }
+        return dp[n - 1];
+    }
+}
 ```
 
 ### **...**
diff --git a/solution/0100-0199/0198.House Robber/README_EN.md b/solution/0100-0199/0198.House Robber/README_EN.md
@@ -41,13 +41,42 @@
 ### **Python3**
 
 ```python
-
+class Solution:
+    def rob(self, nums: List[int]) -> int:
+        if not nums:
+            return 0
+        n = len(nums)
+        if n == 1:
+            return nums[0]
+        dp = [0 for _ in range(n)]
+        dp[0] = nums[0]
+        dp[1] = max(nums[0], nums[1])
+        for i in range(2, n):
+            dp[i] = max(dp[i - 1], dp[i - 2] + nums[i])
+        return dp[n - 1]
 ```
 
 ### **Java**
 
 ```java
-
+class Solution {
+    public int rob(int[] nums) {
+        int n;
+        if (nums == null || (n = nums.length) == 0) {
+            return 0;
+        }
+        if (n == 1) {
+            return nums[0];
+        }
+        int[] dp = new int[n];
+        dp[0] = nums[0];
+        dp[1] = Math.max(nums[0], nums[1]);
+        for (int i = 2; i < n; ++i) {
+            dp[i] = Math.max(dp[i - 1], dp[i - 2] + nums[i]);
+        }
+        return dp[n - 1];
+    }
+}
 ```
 
 ### **...**
diff --git a/solution/0100-0199/0198.House Robber/Solution.java b/solution/0100-0199/0198.House Robber/Solution.java
@@ -1,23 +1,18 @@
 class Solution {
     public int rob(int[] nums) {
-        if (nums == null || nums.length == 0) {
+        int n;
+        if (nums == null || (n = nums.length) == 0) {
             return 0;
         }
-        
-        int n = nums.length;
         if (n == 1) {
             return nums[0];
         }
-        
-        int[] result = new int[n];
-        result[0] = nums[0];
-        result[1] = Math.max(nums[0], nums[1]);
-        
+        int[] dp = new int[n];
+        dp[0] = nums[0];
+        dp[1] = Math.max(nums[0], nums[1]);
         for (int i = 2; i < n; ++i) {
-            result[i] = Math.max(nums[i] + result[i - 2], result[i - 1]);
+            dp[i] = Math.max(dp[i - 1], dp[i - 2] + nums[i]);
         }
-        
-        return result[n - 1];
-        
+        return dp[n - 1];
     }
 }
diff --git a/solution/0100-0199/0198.House Robber/Solution.py b/solution/0100-0199/0198.House Robber/Solution.py
@@ -1,19 +1,13 @@
 class Solution:
-    def rob(self, nums):
-        """
-        :type nums: List[int]
-        :rtype: int
-        """
-        length=len(nums)
-        if length == 0:
+    def rob(self, nums: List[int]) -> int:
+        if not nums:
             return 0
-        if length == 1:
+        n = len(nums)
+        if n == 1:
             return nums[0]
-        res=[0]*length
-        res[0]=nums[0]
-        res[1]=max(nums[0],nums[1])
-        
-        for i in range(2,length):
-            res[i]=max(nums[i]+res[i-2],res[i-1])
-        
-        return res[-1]
+        dp = [0 for _ in range(n)]
+        dp[0] = nums[0]
+        dp[1] = max(nums[0], nums[1])
+        for i in range(2, n):
+            dp[i] = max(dp[i - 1], dp[i - 2] + nums[i])
+        return dp[n - 1]
