LeetCode 474. Ones and Zeroes

最新推荐文章于 2025-09-12 17:31:31 发布

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最新推荐文章于 2025-09-12 17:31:31 发布

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文章标签：数据结构与算法

原文链接：http://www.cnblogs.com/A-Little-Nut/p/10067413.html

本文探讨了如何使用动态规划算法解决给定0和1数量下，从字符串数组中找到最大数量的可形成字符串的问题。通过实例演示了算法的实现过程，以及如何在限制条件下最大化资源利用。

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 m0s and n1s 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 m0s and n1s. 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".

分析

等价于给你一个集合，集合里的是pair, pair.first是string 0的个数，pair.second是string 1的个数。然后给你m,n。让你在集合里尽可能多的选择pair出来，是这些pair的first 之和不大于m, second之和不大于n，就酱。
我采用的是动态规划，dp[i][j]表示的意义是当m=i, n=j时，能拿取的最大pair数，对每个pair有两种状态，其一是拿取，dp[i][j]=dp[i-pair.first][j-pair.second]+1。其二是不拿，dp[i][j]=dp[i][j];
故状态转移方程是dp[i][j]=max ( dp[i-pair.first][j-pair.second]+1, dp[i][j]=dp[i][j])

class Solution {
public:
    int findMaxForm(vector<string>& strs, int m, int n) {
        vector<pair<int,int>> nums;
        for(int i=0;i<strs.size();i++){
            pair<int,int> p={0,0};
            for(int j=0;j<strs[i].size();j++)
                strs[i][j]-'0'==1?p.second++:p.first++;
            nums.push_back(p);
        }
        vector<vector<int>> dp(m+1,vector<int>(n+1,0));
        for(auto o:nums)
            for(int i=m;i>=o.first;i--)
                for(int j=n;j>=o.second;j--)
                    dp[i][j]=max(dp[i][j],dp[i-o.first][j-o.second]+1);
        return dp[m][n];
    }
};

转载于:https://www.cnblogs.com/A-Little-Nut/p/10067413.html