Ones and Zeroes 问题及解法

问题描述：

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:

The given numbers of 

0s

 and 

1s

 will
both not exceed 

100

The size of given string array won't exceed 

600

.

示例:

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”

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".

问题分析：

我们定义一种状态转移数组dp[i][j]，表示i个0和j个1最多能组成几个字串，那么其转移过程就是dp[i][j] = max(dp[i][j], dp[i - zero][j - one] + 1);其中zero和one是当前字串中0

和1的个数。

过程详见代码：

class Solution {
public:
int findMaxForm(vector<string>& strs, int m, int n) {
vector<vector<int>> dp(m + 1, vector<int>(n + 1, 0));
for (auto s : strs)
{
int zero = 0,one = 0;
for (char ch : s)
{
if (ch == '0') zero++;
else one++;
}
for (int i = m; i >= zero; i--)
{
for (int j = n; j >= one; j--)
{
dp[i][j] = max(dp[i][j], dp[i - zero][j - one] + 1);
}
}
}

return dp[m]
;
}
};