Leetcode-474. Ones and Zeroes

前言：为了后续的实习面试，开始疯狂刷题，非常欢迎志同道合的朋友一起交流。因为时间比较紧张，目前的规划是先过一遍，写出能想到的最优算法，第二遍再考虑最优或者较优的方法。如有错误欢迎指正。博主首发CSDN，mcf171专栏。

博客链接：mcf171的博客

——————————————————————————————

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:

1. The given numbers of 0s and 1s will both not exceed 100
2. The size of given string array won't exceed 600.

Example 1:

Input: Array = {"10", "0001", "111001", "1", "0"}, m = 5, n = 3Output: 4Explanation: 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 = 1Output: 2Explanation: You could form "10", but then you'd have nothing left. Better form "0" and "1".

这个题目看完之后其实还是很明白题意，同时也很明白应该是用动态规划来做，但是动态规划一直是我的一个弱项，一直没想明白是怎么迭代的。参考了论坛的解法大致明白了思路，因为每次迭代都会更新，所以无所谓那个字符串先后。Your runtime beats 48.05% of java submissions

public class Solution {    public int findMaxForm(String[] strs, int m, int n) {        int [][] tables = new int[m + 1][n + 1];        for(String item : strs){            int m_0 = 0, n_1 = 0;            for(int i = 0; i < item.length(); i ++){                if(item.charAt(i) == '0') m_0 ++;                else n_1 ++;            }            for(int i = m; i >= m_0; i --)                for(int j =n; j >= n_1; j--){                    tables[i][j] = Math.max(tables[i][j], 1 + tables[i-m_0][j-n_1]);                }        }        return tables[m][n];    }}