LeetCode Ones and Zeroes

Posted 2020-10-09 Dylan_Java_NYC

原题链接在这里：https://leetcode.com/problems/ones-and-zeroes/description/

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:

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 = 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".

题解：

DP问题. 

求m个0和n个1最多能包含几个给出的string. 保存历史信息dp[i][j] 是i个0和j个1最多能包含几个string.

递推时, 若包含当前String s, s中有x个0和y个1. 那么dp[i][j] = 1 + dp[i-x][j-y]. 若不包含s, dp[i][j] = dp[i][j]. 两者取较大值.

初始化都是0.

答案dp[m][n].

Time Complexity: O(strs.length*m*n).

Space: O(1).

AC Java:

 1 class Solution {
 2     public int findMaxForm(String[] strs, int m, int n) {
 3         if(strs == null || strs.length == 0 || m < 0 || n < 0){
 4             return 0;
 5         }
 6         
 7         int [][] dp = new int[m+1][n+1];
 8         for(String s : strs){
 9             int [] count = countZerosOnes(s); 
10             for(int i = m; i>=count[0]; i--){
11                 for(int j = n; j>=count[1]; j--){
12                     dp[i][j] = Math.max(dp[i][j], 1+dp[i-count[0]][j-count[1]]);
13                 }
14             }
15         }
16         return dp[m][n];
17     }
18     
19     private int [] countZerosOnes(String s){
20         int [] res = new int[2];
21         for(int i = 0; i<s.length(); i++){
22             res[s.charAt(i)-‘0‘]++;
23         }
24         return res;
25     }
26 }