474. Ones and Zeroes

最新推荐文章于 2021-03-29 21:50:06 发布

原创最新推荐文章于 2021-03-29 21:50:06 发布 · 216 阅读

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本文探讨了一个计算机科学领域的优化问题——如何利用有限数量的0和1来构造给定数组中的尽可能多的字符串。每个0和1只能使用一次，并且总数不超过100个。通过实例说明了如何求解此问题，并提供了一段C++代码实现。

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 only0s 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 mostonce.
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".

class Solution {  
public:  
    int findMaxForm(vector<string>& strs, int m, int n) {  
        vector<vector<int>> dp(m+1, vector<int>(n+1, 0));  
        int cnt0, cnt1;  
          
        for(auto str:strs) {  
            cnt0 = 0;  
            cnt1 = 0;  
            for(auto c:str) {  
                if(c == '0') ++cnt0;  
                else ++cnt1;  
            }  
              
            for(int i=m; i>=cnt0; --i) {  
                for(int j=n; j>=cnt1; --j) {  
                    dp[i][j] = max(dp[i][j], dp[i-cnt0][j-cnt1]+1);  
                }  
            }  
        }  
          
        return dp[m][n];  
    }  
};