Ones and Zeroes 问题及解法

最新推荐文章于 2020-04-09 17:45:39 发布

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最新推荐文章于 2020-04-09 17:45:39 发布

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分类专栏： LeetCode算法练习（Medium）文章标签： LeetCode Medium DP

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本文探讨了在有限的0和1资源下，如何通过动态规划的方法找出能够组成的字符串的最大数量。介绍了状态转移方程的概念及其应用，并给出了具体的实现代码。

问题描述：

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 0and 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.

示例:

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”

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".

问题分析：

我们定义一种状态转移数组dp[i][j]，表示i个0和j个1最多能组成几个字串，那么其转移过程就是dp[i][j] = max(dp[i][j], dp[i - zero][j - one] + 1);其中zero和one是当前字串中0

和1的个数。

过程详见代码：

class Solution {
public:
    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 zero = 0,one = 0;
			for (char ch : s)
			{
				if (ch == '0') zero++;
				else one++;
			}
			for (int i = m; i >= zero; i--)
			{
				for (int j = n; j >= one; j--)
				{
					dp[i][j] = max(dp[i][j], dp[i - zero][j - one] + 1);
				}
			}
		}
		
		return dp[m][n];
	}
};