资源限制下的字符串组合-优快云博客

本文链接：https://blog.youkuaiyun.com/JB15336039/article/details/73028415

本文探讨了一种在有限资源条件下，即特定数量的0和1下，如何从包含0和1的字符串数组中选取最多数量的字符串的问题。通过动态规划的方法，使用二维数组记录最大组合数量，最终得出在给定资源限制下的最优解。

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474. Ones and Zeroes

Total Accepted: 10055
Total Submissions: 26354
Difficulty: Medium
Contributors:piy9

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 0sand 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".

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思路：

使用二维数组mn来记录组成i个'0'和j个'1'的最大元素个数，因为要求的是m个'0'和n个'1'，所以mn的边界可定为m和n，由此可得mn（i，j）= max（mn（i，j），mn（i - n0，j - n1)，其中n0和n1分别为字符串数组中某个元素的'0'个数和'1'个数，对每个元素进行从n0到m、n1到n的mn的更新，这样最后的mn（m，n）就是所求的结果。

代码：

class Solution {
public:
    int findMaxForm(vector
      
       & strs, int m, int n) {
        int size = strs.size();
        vector
       
        
         > mn(m + 1, vector
         
          (n + 1, 0));
        int n0 = 0;
        int n1 = 0;
        for(int i = 0; i < size; i ++){
            int size1 = strs[i].size();
            for(int j = 0; j < size1; j ++){
                if(strs[i][j] == '0') n0 ++;
                if(strs[i][j] == '1') n1 ++;
            }
            for(int k = m; k >= n0; k --){
                for(int l = n; l >= n1; l --){
                    mn[k][l] = max(mn[k][l], mn[k- n0][l - n1] + 1);
                }
            }
            n0 = n1 = 0;
        }
        return mn[m][n];
    }
};