Merge pull request #2508 from MatthewMaxy/master

添加0279.完全平方数 Python2版本 ; 0139.单词拆分Python DP剪枝方法 ; 背包问题理论基础多重背包 Python解法；添加0583.两个字符串的删除操作 Python解法2

problems/0139.单词拆分.md

@@ -394,7 +394,28 @@ class Solution:
                     dp[j] = dp[j] or (dp[j - len(word)] and word == s[j - len(word):j])
         return dp[len(s)]
 ```
+DP（剪枝）
 
+```python
+class Solution(object):
+    def wordBreak(self, s, wordDict):
+
+        # 先对单词按长度排序
+        wordDict.sort(key=lambda x: len(x))
+        n = len(s)
+        dp = [False] * (n + 1)
+        dp[0] = True
+        # 遍历背包
+        for i in range(1, n + 1):
+            # 遍历单词
+            for word in wordDict:
+                # 简单的 “剪枝”
+                if len(word) > i:
+                    break
+                dp[i] = dp[i] or (dp[i - len(word)] and s[i - len(word): i] == word)
+        return dp[-1]
+
+```
 
 
 ### Go：

problems/0279.完全平方数.md

@@ -271,6 +271,26 @@ class Solution:
         # 返回结果
         return dp[n]
 
+```
+```python
+class Solution(object):
+    def numSquares(self, n):
+        # 先把可以选的数准备好，更好理解
+        nums, num = [], 1
+        while num ** 2 <= n:
+            nums.append(num ** 2)
+            num += 1
+        # dp数组初始化
+        dp = [float('inf')] * (n + 1)
+        dp[0] = 0
+
+        # 遍历准备好的完全平方数
+        for i in range(len(nums)):
+            # 遍历背包容量
+            for j in range(nums[i], n+1):
+                dp[j] = min(dp[j], dp[j-nums[i]]+1)
+        # 返回结果
+        return dp[-1]
         
 
 ```

problems/0583.两个字符串的删除操作.md

@@ -234,6 +234,25 @@ class Solution:
         return dp[-1][-1]
 ```
 
+> 版本 2
+
+```python
+class Solution(object):
+    def minDistance(self, word1, word2):
+        m, n = len(word1), len(word2)
+        
+        # dp 求解两字符串最长公共子序列
+        dp = [[0] * (n+1) for _ in range(m+1)]
+        for i in range(1, m+1):
+            for j in range(1, n+1):
+                if word1[i-1] == word2[j-1]:
+                    dp[i][j] = dp[i-1][j-1] + 1
+                else:
+                    dp[i][j] = max(dp[i-1][j], dp[i][j-1])
+                    
+        # 删去最长公共子序列以外元素
+        return m + n - 2 * dp[-1][-1]
+```
 ### Go：
 
 ```go

problems/背包问题理论基础多重背包.md

@@ -204,6 +204,29 @@ class multi_pack{
 ```
 ### Python：
 
+```python
+
+C, N = input().split(" ")
+C, N = int(C), int(N)
+
+# value数组需要判断一下非空不然过不了
+weights = [int(x) for x in input().split(" ")]
+values = [int(x) for x in input().split(" ") if x]
+nums = [int(x) for x in input().split(" ")]
+
+dp = [0] * (C + 1)
+# 遍历背包容量
+for i in range(N):
+    for j in range(C, weights[i] - 1, -1):
+        for k in range(1, nums[i] + 1):
+            # 遍历 k，如果已经大于背包容量直接跳出循环
+            if k * weights[i] > j:
+                break
+            dp[j] = max(dp[j], dp[j - weights[i] * k] + values[i] * k) 
+print(dp[-1])
+
+```
+
 ### Go：