474. Ones and Zeroes

474. Ones and Zeroes

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• Total Accepted: 9812
• Total Submissions: 25753
• Difficulty: Medium
• Contributors:piy9

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 0sand 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 mostonce.

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".

题目：给一个string的数组，并且给两个整数m和n，取一个string的组合，使得组合中0的个数为m，1的个数为n。找出一个组合，使其中的字符串个数最多。

提醒：这题的结果一定存在，就是说一定有符合 0s = m, 1s = n的组合，不需要考虑存在性。

这一题求的是最优结果，我们可以尝试使用动态规划。

我们另 f(a, b)为0的个数为a，1的个数为b的最优组合（字符串个数最多）

那么有递推式：

f(a, b) = max(f(a, b), f(a-i, b-j) + 1)

其中， i是其中一个字符串s中0的个数，j是s中1的个数。

这个递推式的意思就是，是在求f(a, b)时否选择字符串s加入组合中

代码如下：

int findMaxForm(vector<string>& strs, int m, int n) {
      int f[m+3][n+3];
      for (int i = 0; i <= m; i++)
          for (int j = 0; j <= n; j++)
              f[i][j] = 0;                         //全部初始化为0
    
    for (string &s: strs){                //遍历vector
    int one = 0, zero = 0;
    for (char c: s)
    if (c == '0') zero++;
    else one++;                        //计算字符串s中0的个数和1的个数
    for (int i = m; i >= zero; i--)       //更新f二维数组，使其达到此时最优
       for (int j = n; j >= one; j--)   //如果i<zero或者j<one，则i-zero 和j-one为负数，不存在
       f[i][j] = max(f[i][j], f[i-zero][j-one]+1);   
}
return f[m][n];

}