474. Ones and Zeroes

题目

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 0s 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 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".

题意

给m 个0 和 n个1, 用这些01构造给定数组中的字符串, 求最多能构造出的字符串是多少个

分析

参考了这里

// dp[i][j] = the max number of strings that can be formed with i 0's and j 1's    // from the first few strings up to the current string s    // Catch: have to go from bottom right to top left    // Why? If a cell in the dp is updated(because s is selected),    // we should be adding 1 to dp[i][j] from the previous iteration (when we were not considering s)    // If we go from top left to bottom right, we would be using results from this iteration => overcounting循环的方式根据状态转移方程,否则会重复计算

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));        for (auto &s : strs) {            // count the num of 0 & 1 in string s            int count0 = 0, count1 = 0;            for (auto c : s)                if (c == '0') count0++;            count1 = s.size() - count0;            // DP part            for (int i = m; i >= count0; i--) {                for (int j = n; j >= count1; j--) {                    dp[i][j] = max(dp[i][j], dp[i-count0][j-count1]+1);                }            }        }        return dp[m][n];    }};