[LeetCode]Ones and Zeroes

题目描述：

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 m0s and n1s 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 m0s and n1s. Each 0 and 1 can be used at most once.

Note:

The given numbers of 0s and 1s will both not exceed 100
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".

题目大意：

给定一组01字符串strs。求用m个0和n个1最多可以组成多少个strs中的字符串。

解题思路：

动态规划（Dynamic Programming）

二维01背包问题（Knapsack Problem）

状态转移方程：

for s in strs:
    zero, one = s.count('0'), s.count('1')
    for x in range(m, zero - 1, -1):
        for y in range(n, one - 1, -1):
            dp[x][y] = max(dp[x - zero][y - one] + 1, dp[x][y])

上式中，dp[x][y]表示至多使用x个0，y个1可以组成字符串的最大数目

Python代码：

class Solution(object):
    def findMaxForm(self, strs, m, n):
        """
        :type strs: List[str]
        :type m: int
        :type n: int
        :rtype: int
        """
        dp = [[0] * (n + 1) for x in range(m + 1)]
        for s in strs:
            zero, one = s.count('0'), s.count('1')
            for x in range(m, zero - 1, -1):
                for y in range(n, one - 1, -1):
                    dp[x][y] = max(dp[x - zero][y - one] + 1, dp[x][y])
        return dp[m][n]

Java代码：

public class Solution {
    public int findMaxForm(String[] strs, int m, int n) {
        int dp[][] = new int[m + 1][n + 1];
        int ans = dp[0][0] = 0;
        for (String s : strs) {
            int zero = 0, one = 0;
            for (int i = 0; i < s.length(); i++) {
                if (s.charAt(i) == '0') {
                    zero++;
                } else {
                    one++;
                }
            }
            for (int i = m; i > zero - 1; i--) {
                for (int j = n; j > one - 1; j--) {
                    dp[i][j] = Math.max(dp[i][j], dp[i - zero][j - one] + 1);
                }
            }
        }
        return dp[m][n];
    }
}