LeetCode474 - Ones and Zeros - Medium (Python)

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In the computer world, use restricted resource you have to generate maximum benefit is what we always want to pursue.

For now, suppose you are a dominator of m 0s and n 1s respectively. On the other hand, there is an array with strings consisting of only 0s and 1s.

Now your task is to find the maximum number of strings that you can form with given m 0s and n 1s. Each 0 and 1 can be used at most once.

Note:

1. The given numbers of 0s and 1s will both not exceed 100
2. The size of given string array won‘t exceed 600.

Example 1:

Input: Array = {"10", "0001", "111001", "1", "0"}, m = 5, n = 3
Output: 4

Explanation: This are totally 4 strings can be formed by the using of 5 0s and 3 1s, which are “10,”0001”,”1”,”0”

Example 2:

Input: Array = {"10", "0", "1"}, m = 1, n = 1
Output: 2

Explanation: You could form "10", but then you‘d have nothing left. Better form "0" and "1".

思路：dp
dp[i][j]代表i个0和j个1可以形成的最多的字符数量。dp[i][j]  = max(1+dp[i-c[‘0‘]][j-c[‘1‘]], dp[i][j]) 因为dp有两种情况，选这个字符以及不选这个字符，选这个字符的话，dp[i][j] = 1+

dp[i-c[‘0‘]][j-c[‘1‘]], 不选的话dp[i][j] = dp[i][j]. 
最后返回dp[-1][-1] 

class Solution:
    def findMaxForm(self, strs: List[str], m: int, n: int) -> int:
        if not strs or len(strs) == 0:
            return False
        dp = [[0 for _ in range(n+1)] for _ in range(m+1)]
        
        for s in strs:
            c = collections.Counter(s)
            
            
            for zero in range(m, c[‘0‘]-1, -1):
                for one in range(n, c[‘1‘]-1, -1):
                    dp[zero][one] = max(1+dp[zero-c[‘0‘]][one-c[‘1‘]], dp[zero][one])
           # print(dp)
        return dp[-1][-1]

LeetCode474 - Ones and Zeros - Medium (Python)

标签：nat   div   ber   this   false   array   turn   list   思路   

原文地址：https://www.cnblogs.com/sky37/p/12244742.html