原題地址

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

題意

給定一個(gè)包含01串的vector，同時(shí)給定可用的0和1的數(shù)目，在可用數(shù)目范圍內(nèi)從vector中挑選01串，求最多能從vector中挑出多少01串。

思路

背包問(wèn)題的變式吧。
遞推方程為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個(gè)串，給定j個(gè)1，k個(gè)0的情況下能拿出的串的數(shù)目。（代碼里i、j、k都是從1開(kāi)始算的）

代碼

1

提交以下代碼得到了runtime error，看了下失敗的例子，應(yīng)該是01串的數(shù)目太多了，所以dp數(shù)組太大爆掉了

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 ];
    }
};

所以優(yōu)化以下空間復(fù)雜度，去掉表示字符串?dāng)?shù)目的那一維

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 ];
    }
};