LeetCode #474 - Ones and Zeroes

题目描述：

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".

给定一个字符串数组和固定个数的“0”和“1”，求问用这些“0”和“1”可以组成数组中的那些字符串，也就是这些字符串的“0”和“1”的总数不超过给定的数目。采用动态规划，用dp[i][j]表示用i个“0”和j个“1”可以构造的字符串数目，那么遍历每一个字符串，确定它包含的“0”和"1"的数目，对于dp[i][j]有递推关系dp[i][j]=max(dp[i][j],dp[i-zeros][j-ones]+1)。需要注意的是每次i都是从m开始遍历，j都是从n开始遍历，目的就是为了让上一轮的dp的结果来更新当前的dp数组，否则在一轮循环中如果strs[k]已经用来更新了dp[a][b]，然后又可能用更新后的dp[a][b]来更新dp[c][d]，导致strs[k]使用了两次，这明显是错误的，而从大往小遍历就必然用到的是前一轮对应的数组值。

class Solution {
public:
    int findMaxForm(vector<string>& strs, int m, int n) {
        vector<vector<int>> dp(m+1,vector<int>(n+1,0));
        for(int k=0;k<strs.size();k++)
        {
            int zeros=get_count(strs[k]).first;
            int ones=get_count(strs[k]).second;
            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];
    }
    
    pair<int,int> get_count(string s)
    {
        int a=0;
        int b=0;
        for(int i=0;i<s.size();i++)
        {
            if(s[i]=='0') a++;
            else b++;
        }
        return pair<int,int>(a,b);
    }
};

LeetCode #474 - Ones and Zeroes

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