[leetcode] 474. Ones and Zeroes

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 only0s 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 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".

这道题是从字符串集合中尽可能多的选出字符串并保证0和1个数不超过给定值，题目难度为Medium。

题目和0-1背包问题大同小异，区别是这里限制0和1个数，而0-1背包问题限制总重量，算是动态规划的经典题目，不了解0-1背包问题的同学可以问下度娘。

这里用dp[i][j][k]表示前i个字符串在0个数不超过j、1个数不超过k时最多能选取的字符串个数。统计第i个字符串中0和1个数分别为cnt0和cnt1，如果取第i个字符串则dp[i][j][k] = dp[i-1][j-cnt0][k-cnt1] + 1，如果不取第i个字符串则dp[i][j][k] = dp[i-1][j][k]，取两者大的作为dp[i][j][k]的值。由于dp[i][j][k]只与dp[i-1][*][*]相关，所以这里可以重复使用m*n个数据将空间复杂度降为O(m*n)，只需在遍历时从后向前遍历即可。具体代码：

class Solution {public:    int findMaxForm(vector<string>& strs, int m, int n) {        vector<vector<int>> dp(m+1, vector<int>(n+1, 0));        int cnt0, cnt1;                for(auto str:strs) {            cnt0 = 0;            cnt1 = 0;            for(auto c:str) {                if(c == '0') ++cnt0;                else ++cnt1;            }                        for(int i=m; i>=cnt0; --i) {                for(int j=n; j>=cnt1; --j) {                    dp[i][j] = max(dp[i][j], dp[i-cnt0][j-cnt1]+1);                }            }        }                return dp[m][n];    }};