# # 【LeetCode】474. Ones and Zeroes 解题报告（Python）

## # 题目描述：

n 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".
``````

## # 解题方法

``````class Solution(object):
def findMaxForm(self, strs, m, n):
"""
:type strs: List[str]
:type m: int
:type n: int
:rtype: int
"""
# m个0, n个1能组成的最长字符串
dp = [[0 for _ in range(n + 1)] for _ in range(m + 1)]
for str in strs:
zeros, ones = 0, 0
for c in str:
if c == "0":
zeros += 1
elif c == "1":
ones += 1
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[m][n]
``````

http://www.cnblogs.com/grandyang/p/6188893.html https://kingsfish.github.io/2017/07/23/Leetcode-474-Ones-and-Zeros/

## # 日期

2018 年 9 月 23 日 —— 今天是实验室第一个打卡的