[Leetcode] 474. Ones and Zeroes 解题报告

题目：

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

.

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：

又是动态规划的一个变种。我们定义dp[i][j]表示从第0个字符串到截止当前字符串，用i个0和j个1可以构成的字符串的最大数目。那么状态转移方程是：dp[i][j] = max(dp[i - zero_num][j - one_num])，其中zero_num和one_num分别是当前字符串中0和1的数目，i <= zero_num <= m, j <= one_num <= n。需要注意的是：我们必须从后往前更新，这是因为如果从前往后更新，那么如果更新了dp[i][j]，就意味着当前dp[i][j]的数量已经包含了s，那么在计算dp[i
+ ...][j + ...]的时候，用到的dp[i][j]是已经被更新过的（也就是包含了s的数量），所以就会导致重复计算。

算法的时间复杂度是O(l * m * n)，其中l是字符串的数量。算法的空间复杂度是O(m*n)。

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 numZeroes = 0, numOnes = 0;
for (auto c : s) {
if (c == '0')
numZeroes++;
else if (c == '1')
numOnes++;
}
for (int i = m; i >= numZeroes; i--) {
for (int j = n; j >= numOnes; j--) {
dp[i][j] = max(dp[i][j], dp[i - numZeroes][j - numOnes] + 1);
}
}
}
return dp[m]
;
}
};