原题地址

https://leetcode.com/problems/ones-and-zeroes/description/

题意

给定一个包含01串的vector，同时给定可用的0和1的数目，在可用数目范围内从vector中挑选01串，求最多能从vector中挑出多少01串。

思路

背包问题的变式吧。
递推方程为dp[i][j][k] = max(dp[ i-1 ][ j ][ k ], dp[ i-1 ][ j-count1(str[i]) ][ k-count0(str[i]) ])
dp[i][j][k]表示：前i个串，给定j个1，k个0的情况下能拿出的串的数目。（代码里i、j、k都是从1开始算的）

代码

1

提交以下代码得到了runtime error，看了下失败的例子，应该是01串的数目太多了，所以dp数组太大爆掉了

class Solution {
public:
    int count0(string s) {
        int count = 0;
        for (int i = 0; i < s.size(); i++) {
            if (s[i] == '0') count++;
        }
        return count;
    }

    int count1(string s) {
        int count = 0;
        for (int i = 0; i < s.size(); i++) {
            if (s[i] == '1') count++;
        }
        return count;
    }

    int findMaxForm(vector<string>& strs, int m, int n) {

        int s = strs.size();
        int dp[s + 1][m + 1][n + 1];

        for (int j = 0; j <= m; j++) {
            for (int k = 0; k <= n; k++) {
                dp[0][j][k] = 0;
            }
        }


        for (int i = 1; i <= s; i++) {
            for (int j = 0; j <= m; j++) {
                for (int k = 0; k <= n; k++) {
            
                    int temp = 0;
                    if (j  >= count0(strs[i-1])  && k  >= count1(strs[i-1]) ) {
                        temp = dp[i - 1][j - count0(strs[i-1])][k  - count1(strs[i-1])] + 1;
                    }
                    dp[i][j][k] = max(dp[i - 1][j][k], temp);
                }
            }
        }
        return dp[s ][m ][n ];
    }
};

所以优化以下空间复杂度，去掉表示字符串数目的那一维

2

class Solution {
public:
    int count0(string s) {
        int count = 0;
        for (int i = 0; i < s.size(); i++) {
            if (s[i] == '0') count++;
        }
        return count;
    }

    int count1(string s) {
        int count = 0;
        for (int i = 0; i < s.size(); i++) {
            if (s[i] == '1') count++;
        }
        return count;
    }

    int findMaxForm(vector<string>& strs, int m, int n) {
        int s = strs.size();
        int dp[m + 1][n + 1];

        for (int j = 0; j <= m; j++) {
            for (int k = 0; k <= n; k++) {
                dp[j][k] = 0;
            }
        }
        for (int i = 1; i <= s; i++) {
            for (int j = m; j >= 0; j--) {
                for (int k = n; k >= 0; k--) {
                    int temp = 0;
                    if (j  >= count0(strs[i - 1])  && k  >= count1(strs[i - 1]) ) {
                        temp = dp[j - count0(strs[i - 1])][k  - count1(strs[i - 1])] + 1;
                    }
                    dp[j][k] = max(dp[j][k], temp);
                }
            }
        }
        return dp[m ][n ];
    }
};