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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
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@ -394,7 +394,28 @@ class Solution:
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dp[j] = dp[j] or (dp[j - len(word)] and word == s[j - len(word):j])
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return dp[len(s)]
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```
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DP(剪枝)
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```python
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class Solution(object):
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def wordBreak(self, s, wordDict):
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# 先对单词按长度排序
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wordDict.sort(key=lambda x: len(x))
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n = len(s)
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dp = [False] * (n + 1)
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dp[0] = True
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# 遍历背包
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for i in range(1, n + 1):
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# 遍历单词
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for word in wordDict:
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# 简单的 “剪枝”
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if len(word) > i:
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break
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dp[i] = dp[i] or (dp[i - len(word)] and s[i - len(word): i] == word)
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return dp[-1]
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```
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### Go:
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@ -271,7 +271,27 @@ class Solution:
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# 返回结果
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return dp[n]
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```
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```python
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class Solution(object):
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def numSquares(self, n):
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# 先把可以选的数准备好,更好理解
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nums, num = [], 1
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while num ** 2 <= n:
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nums.append(num ** 2)
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num += 1
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# dp数组初始化
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dp = [float('inf')] * (n + 1)
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dp[0] = 0
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# 遍历准备好的完全平方数
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for i in range(len(nums)):
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# 遍历背包容量
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for j in range(nums[i], n+1):
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dp[j] = min(dp[j], dp[j-nums[i]]+1)
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# 返回结果
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return dp[-1]
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```
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### Go:
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@ -234,6 +234,25 @@ class Solution:
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return dp[-1][-1]
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```
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> 版本 2
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```python
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class Solution(object):
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def minDistance(self, word1, word2):
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m, n = len(word1), len(word2)
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# dp 求解两字符串最长公共子序列
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dp = [[0] * (n+1) for _ in range(m+1)]
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for i in range(1, m+1):
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for j in range(1, n+1):
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if word1[i-1] == word2[j-1]:
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dp[i][j] = dp[i-1][j-1] + 1
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else:
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dp[i][j] = max(dp[i-1][j], dp[i][j-1])
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# 删去最长公共子序列以外元素
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return m + n - 2 * dp[-1][-1]
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```
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### Go:
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```go
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@ -204,6 +204,29 @@ class multi_pack{
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```
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### Python:
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```python
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C, N = input().split(" ")
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C, N = int(C), int(N)
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# value数组需要判断一下非空不然过不了
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weights = [int(x) for x in input().split(" ")]
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values = [int(x) for x in input().split(" ") if x]
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nums = [int(x) for x in input().split(" ")]
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dp = [0] * (C + 1)
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# 遍历背包容量
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for i in range(N):
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for j in range(C, weights[i] - 1, -1):
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for k in range(1, nums[i] + 1):
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# 遍历 k,如果已经大于背包容量直接跳出循环
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if k * weights[i] > j:
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break
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dp[j] = max(dp[j], dp[j - weights[i] * k] + values[i] * k)
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print(dp[-1])
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```
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### Go:
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