【leetcode】474. Ones and Zeroes

题目如下：

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".

解题思路：二维背包问题。记dp[i][j][k]为前i个元素中使用了j个0，k个1可以获得的最大值。对于Array[i]来说，有选和不选两种操作，如果不选，那么dp[i][j][k] = dp[i-1][j][k]，如果选那么则有dp[i][j][k] = dp[i-1][j-i0][k-i1] (i0和i1分别为Array[i]中0和1的数量)。这种解法的时间复杂度是O(n^3)，用python会超时，用C++则能通过。

代码如下：

python

### TEL, C++ Accpeted

class Solution(object):
    def findMaxForm(self, strs, m, n):
        """
        :type strs: List[str]
        :type m: int
        :type n: int
        :rtype: int
        """
        res = 0
        dp = [[[0 for x in range(n + 1)] for x in range(m + 1)] for x in strs]
        c1 = strs[0].count('1')
        c0 = strs[0].count('0')
        if c0 <= m and c1 <= n:
            dp[0][c0][c1] = 1
            res = max(res, dp[0][c0][c1])
        count = 0
        for i in range(1,len(strs)):
            for j in range(m + 1):
                for k in range(n + 1):
                    count += 1
                    dp[i][j][k] = max(dp[i][j][k],dp[i-1][j][k])
                    c1 = strs[i].count('1')
                    c0 = strs[i].count('0')
                    if j - c0 >= 0 and k - c1 >= 0:
                        dp[i][j][k] = max(dp[i][j][k], dp[i - 1][j - c0][k - c1] + 1)
                    res = max(res,dp[i][j][k])
        return res

C++

#include <vector>
#include <map>
#include <string>
using namespace std;

class Solution {
public:
    int getCount(string str,char val){
        int count = 0;
        for(int i = 0;i<str.size();i++){
            if (val == str[i]){
                count += 1;
            }
        }
        return count;
    }
    #define MAX(a,b) ((a) < (b) ? (b) : (a))
    int findMaxForm(vector<string>& strs, int m, int n) {
        //memset(dp,0,1000*101*101);
        int dp[600][101][101] = {0};
        int res = 0;
        int c1 = getCount(strs[0],'1');
        int c0 = getCount(strs[0],'0');
        if (c0 <= m && c1 <= n){
            dp[0][c0][c1] = 1;
            res = MAX(res, dp[0][c0][c1]);
        }
        for(int i = 1;i<strs.size();i++){
            for(int j = 0;j<=m;j++){
                for(int k = 0;k<=n;k++){
                    dp[i][j][k] = MAX(dp[i][j][k],dp[i-1][j][k]);
                    c1 = getCount(strs[i],'1');
                    c0 = getCount(strs[i],'0');
                    if (j - c0 >= 0 && k - c1 >= 0){
                        dp[i][j][k] = MAX(dp[i][j][k], dp[i - 1][j - c0][k - c1] + 1);
                    }
                    res = MAX(res,dp[i][j][k]);
                }
            }
        }
        return res;  
    }
};