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474. Ones and Zeroes

2024-09-06 09:47:31   215人阅读

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.

代码如下：（动态规划）

解析：

状态转换公式：dp[i][j]=Math.max(dp[i-a[0]][j-a[1]+1],dp[i][j])

public class Solution {
    public int findMaxForm(String[] strs, int m, int n) {
        if(m==0&&n==0)
        return 0;                                                                                                                           
        if(m<0||n<0)
        return 0;
        int[][] dp=new int[m+1][n+1];
        
        for(int i=0;i<strs.length;i++)
        {
            String s=strs[i];
            int[] a=abs(s);
            for(int j=m;j>=a[0];j--)//备忘录的自顶向下
            {
                for(int k=n;k>=a[1];k--)
                {
                    dp[j][k]=Math.max(dp[j-a[0]][k-a[1]]+1,dp[j][k]);
                }
            }
        }
        return dp[m][n];
    }
    int[] abs(String s){   //用于统计每个字符串中0和1的个数
        int[] a=new int[2];
         for(int i=0;i<s.length();i++)
            {
                if(s.charAt(i)==‘0‘)
                a[0]++;
                else
                a[1]++;
            }
            return a;
    }
}

　　

474. Ones and Zeroes


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