diff --git a/problems/0474.一和零.md b/problems/0474.一和零.md index 6a178a25..145f64aa 100644 --- a/problems/0474.一和零.md +++ b/problems/0474.一和零.md @@ -210,19 +210,35 @@ class Solution { ``` ### Python +DP(版本一) ```python class Solution: def findMaxForm(self, strs: List[str], m: int, n: int) -> int: - dp = [[0] * (n + 1) for _ in range(m + 1)] # 默认初始化0 + dp = [[0] * (n + 1) for _ in range(m + 1)] # 创建二维动态规划数组,初始化为0 + for s in strs: # 遍历物品 + zeroNum = s.count('0') # 统计0的个数 + oneNum = len(s) - zeroNum # 统计1的个数 + for i in range(m, zeroNum - 1, -1): # 遍历背包容量且从后向前遍历 + for j in range(n, oneNum - 1, -1): + dp[i][j] = max(dp[i][j], dp[i - zeroNum][j - oneNum] + 1) # 状态转移方程 + return dp[m][n] + +``` +DP(版本二) +```python +class Solution: + def findMaxForm(self, strs: List[str], m: int, n: int) -> int: + dp = [[0] * (n + 1) for _ in range(m + 1)] # 创建二维动态规划数组,初始化为0 # 遍历物品 - for str in strs: - ones = str.count('1') - zeros = str.count('0') - # 遍历背包容量且从后向前遍历! + for s in strs: + ones = s.count('1') # 统计字符串中1的个数 + zeros = s.count('0') # 统计字符串中0的个数 + # 遍历背包容量且从后向前遍历 for i in range(m, zeros - 1, -1): for j in range(n, ones - 1, -1): - dp[i][j] = max(dp[i][j], dp[i - zeros][j - ones] + 1) + dp[i][j] = max(dp[i][j], dp[i - zeros][j - ones] + 1) # 状态转移方程 return dp[m][n] + ``` ### Go