474. Ones and Zeroes

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最新推荐文章于 2025-08-10 20:50:18 发布

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分类专栏： LeetCode 文章标签： leetcode 动态规划

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本文介绍了一个计算机科学问题：如何从包含0和1的字符串数组中找出最多数量的字符串，使得这些字符串总共包含指定数量的0和1。文章提供了一种使用动态规划解决此问题的方法，并给出了具体的代码实现。

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474. Ones and Zeroes

题目

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”.

题目要求一个字符串vector中能找够m个0，n个1的最多的字符串数。
用dp[i][j]表示，状态转移方程：

dp[i][j] = max(dp[i][j], dp[i - zeros][j - ones] + 1)

代码实现

class Solution {
public:
    int findMaxForm(vector<string>& strs, int m, int n) {
        vector<vector<int>> dp(m + 1, vector<int>(n + 1, 0));
        for (string str : strs) {
            int zeros = 0, ones = 0;
            for (char c : str) (c == '0') ? ++zeros : ++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];
    }
};