leetcode 474 Ones and Zeroes

最新推荐文章于 2021-07-26 00:23:30 发布

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本文探讨了在有限的0和1资源下，如何从由0和1组成的字符串数组中，找出能使用这些资源形成的最大数量的字符串。通过详细的例子和算法解释，展示了背包问题的一种解决方案，使用三维和二维数组实现动态规划。

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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".

题解：

这是一个背包问题

三维数组如下：

//Make a 3D Vector
typedef vector<int> v1d;
typedef vector<v1d> v2d;
typedef vector<v2d> v3d;

class Solution {
public:
    int findMaxForm(vector<string>& strs, int m, int n) {
        int l = strs.size();
        v3d dp(l+1, v2d(m+1, v1d(n+1, 0))); //dp[l+1][m+1][n+1]
        
        for(int i=1;i<=l;i++)
        {
            string s = strs[i-1];
            int ones = count(s.begin(), s.end(), '1');
            int zeros = s.size()-ones;
            
            for (int j=0;j<=m;j++)
            {
                for (int k=0;k<=n;k++)
                {
                    if (j>=zeros && k>=ones)
                        dp[i][j][k] = max(dp[i-1][j][k], 1 + dp[i-1][j-zeros][k-ones] );
                    else
                        dp[i][j][k] = dp[i-1][j][k];
                    
                }//k
            }//j
        }//i
        
        return dp[l][m][n];
    }
};

二维数组：

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 ones = count(s.begin(), s.end(), '1');
        int zeros= s.size()-ones;
        for (int i=m; i>=zeros; i--) 
            for (int j=n; j>=ones; j--)
                dp[i][j] = max(dp[i][j], dp[i-zeros][j-ones]+1);
    }
    return dp[m][n];
}