[leetcode] 474. Ones and Zeroes

[leetcode] 474. Ones and Zeroes

Description

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 100The 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

题目的意思是：给你m个0，n个1，问能最多组成给定的数的数量。

建立一个二维的DP数组，其中dp[i][j]表示有i个0和j个1时能组成的最多字符串的个数，而对于当前遍历到的字符串，我们统计出其中0和1的个数为zeros和ones，然后dp[i - zeros][j - ones]表示当前的i和j减去zeros和ones之前能拼成字符串的个数，那么加上当前的zeros和ones就是当前dp[i][j]可以达到的个数，我们跟其原有数值对比取较大值即可，所以递推式如下：dp[i][j] = max(dp[i][j], dp[i - zeros][j - ones] + 1);

代码

class Solution {public: int findMaxForm(vector& strs, int m, int n) { vector> dp(m+1,vector(n+1,0)); for(auto str:strs){ int zeros=0; int ones=0; for(auto ch:str){ if(ch=='1'){ ones++; }else{ zeros++; } } 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]; }};

参考文献

​​[LeetCode] Ones and Zeroes 一和零​​

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