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Update 0474.一和零.md
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@ -210,19 +210,35 @@ class Solution {
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```
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### Python
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DP(版本一)
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```python
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class Solution:
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def findMaxForm(self, strs: List[str], m: int, n: int) -> int:
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dp = [[0] * (n + 1) for _ in range(m + 1)] # 默认初始化0
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dp = [[0] * (n + 1) for _ in range(m + 1)] # 创建二维动态规划数组,初始化为0
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for s in strs: # 遍历物品
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zeroNum = s.count('0') # 统计0的个数
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oneNum = len(s) - zeroNum # 统计1的个数
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for i in range(m, zeroNum - 1, -1): # 遍历背包容量且从后向前遍历
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for j in range(n, oneNum - 1, -1):
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dp[i][j] = max(dp[i][j], dp[i - zeroNum][j - oneNum] + 1) # 状态转移方程
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return dp[m][n]
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```
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DP(版本二)
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```python
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class Solution:
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def findMaxForm(self, strs: List[str], m: int, n: int) -> int:
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dp = [[0] * (n + 1) for _ in range(m + 1)] # 创建二维动态规划数组,初始化为0
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# 遍历物品
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for str in strs:
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ones = str.count('1')
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zeros = str.count('0')
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# 遍历背包容量且从后向前遍历!
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for s in strs:
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ones = s.count('1') # 统计字符串中1的个数
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zeros = s.count('0') # 统计字符串中0的个数
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# 遍历背包容量且从后向前遍历
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for i in range(m, zeros - 1, -1):
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for j in range(n, ones - 1, -1):
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dp[i][j] = max(dp[i][j], dp[i - zeros][j - ones] + 1)
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dp[i][j] = max(dp[i][j], dp[i - zeros][j - ones] + 1) # 状态转移方程
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return dp[m][n]
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```
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### Go
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