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 most once.
思路：使用动态规划思想。子问题为有更少的0和更少的1时能组合出来的最大个数。用deep[i][j]表示有i个0和j个1的情况下最大的个数，循环遍历strs的每一个元素，每遍历一次就看当前情况下可以有的最大的组合个数。所以deep[m][n]为想要的最终结果。
class Solution { public: int findMaxForm(vector<string>& strs, int m, int n) { int deep[m + 1][n + 1], num0 = 0, num1 = 0; string temp; for (int i = 0; i < m + 1; i++) { for (int j = 0; j < n + 1; j++) { deep[i][j] = 0; } } for (int i = 0; i < strs.size(); i++) { temp = strs[i]; num0 = 0; num1 = 0; for (int j = 0; j < temp.length(); j++) { if (temp[j] == '0') { num0++; } else { num1++; } } for (int j = m; j >= num0; j--) { for (int k = n; k >= num1; k--) { deep[j][k] = max(deep[j][k], deep[j - num0][k - num1] + 1); } } } return deep[m][n]; } };
