让我们首先声明我们的初始字符串并计算其长度并将它们传递给deleteSubstr(str,length)函数。
string str = "01010110011"; int length = str.length(); cout <<"子串删除次数"<< deleteSubstr(str, length);
在deleteSubstr(string str, int length)函数内部,for 循环运行直到 I 小于长度并分别在遇到 0 和 1 时递增 count_0 和 count_1 变量。然后该函数返回count_0 和count_1 的最小值。
int deleteSubstr(string str, int length){ int count_0 = 0, count_1 = 0; for (int i = 0; i < length; i++) { if (str[i] == '0') count_0++; else count_1++; } return min(count_0, count_1); }
让我们看看以下删除二进制字符串中的“01”或“10”以使其摆脱“01”或“10”的实现 -
#include <iostream> using namespace std; int deleteSubstr(string str, int length){ int count_0 = 0, count_1 = 0; for (int i = 0; i < length; i++) { if (str[i] == '0') count_0++; else count_1++; } return min(count_0, count_1); } int main(){ string str = "01010110011"; int length = str.length(); cout <<"子串删除次数 "<< deleteSubstr(str, length); return 0; }输出结果
上面的代码将产生以下输出 -
子串删除次数 5