动态规划--LC474.一和零

class Solution(object):
    def findMaxForm(self, strs, m, n):
        """
        :type strs: List[str]
        :type m: int
        :type n: int
        :rtype: int
        """
        # 这个题是一个01背包问题，不一样的是，背包的容量是2维的，有m和n
        # 确定dp
        # dp[i][j][k],使用前i个字符串，最多有j个0和k个1的strs的最大子集的大小为dp[i][j][k]
        # i为可以选择的字符串，0<=i<=len(nums),0<=j<=m 0<=k<=n
        # 状态转移方程
        # 1.当j<zeros or k<ones此时不能取字符串nums[i],dp[i][j][k]=dp[i-1][j][k]
        # 2.j>=zeros and k>=ones，此时可以取字符串nums[i]，也可以不取
        # 不取：dp[i][j][k]=dp[i-1][j][k]
        # 取：dp[i][j][k]=dp[i-1][j-zeros][k-ones]+1, +1代表选取当前字符串，子集大小+1
        # 初始化
        dp = [[[0]*(n+1) for _ in range(m+1)] for _ in range(len(strs)+1)]
        # 先遍历元素
        for i in range(1, len(strs)+1):
            # 选取元素后，先计算元素中0和1的数量
            str_ = strs[i-1]
            zeros = str_.count('0')
            ones = str_.count('1')
            for j in range(m+1):
                for k in range(n+1):
                    dp[i][j][k] = dp[i-1][j][k]
                    if j>=zeros and k>=ones:
                        dp[i][j][k] = max(dp[i-1][j][k],dp[i-1][j-zeros][k-ones]+1)
        return dp[-1][-1][-1]

class Solution(object):
    def findMaxForm(self, strs, m, n):
        """
        :type strs: List[str]
        :type m: int
        :type n: int
        :rtype: int
        """
        # 优化，滚动数组
        dp = [[0]*(n+1) for _ in range(m+1)]
        for str_ in strs:
            zeros = str_.count('0')
            ones = str_.count('1')
            # 倒序遍历
            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)
        return dp[-1][-1]