Leetcode 474 Ones and Zeroes

Description

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:
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”

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".

题解

题意很明显，给出一些字符串，和数字m、n，求如果要得出m个0和n个1最多需要多少个字符串。直接用动态规划，用二维数组dp[i][j]表示i个1，j个0的需要多少个字符串，然后核心算法是max(dp[i][j], dp[i - zeroNums][j - oneNums] + 1)，表示拿不取当前字符串和取当前字符串的最大值，然后遍历所有的字符串最后输出结果即可，代码如下:

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 &str : strs) {            int oneNums = 0, zeroNums = 0;            for (auto &c : str) {                if (c == '0')                    zeroNums++;                else                    oneNums++;            }            for (int i = m; i >= zeroNums; i--) {                for (int j = n; j >= oneNums; j--) {                    dp[i][j] = max(dp[i][j], dp[i - zeroNums][j - oneNums] + 1);                }            }        }        return dp[m][n];    }};