C ++中的串联二进制字符串中的最大连续零
假设我们有一个长度为n的二进制字符串，另一个值是k。我们必须将二进制字符串连接k次。然后，我们必须在级联字符串中找到最大连续0个数。假设二进制字符串是“0010010”，并且k=2，那么在将字符串连接了k次之后，它将是“00100100010010”。因此，连续0的最大数量为3。
该方法很简单。如果数字全为0，则答案为n*k。如果字符串包含1，则结果将是字符串的子字符串的最大长度（包含全0），或者是仅包含0的字符串的最大前缀长度与最大后缀的长度之和。仅包含0的字符串。
算法
max_zero_count(str,n,k)-
Begin total := 0 len := 0 for i in range 0 to n, do if str[i] = 0, then increase len else len := 0 total := maximum of total and len done if total = n, then return n * k prefix := length of maximal prefix with only 0 suffix:= length of maximal suffix with only 0 if k > 1, then total := max of total, and (prefix + suffix) return total End
示例
#include <iostream>
using namespace std;
int max_length_substring(string str, int n, int k) {
int total_len = 0;
int len = 0;
for (int i = 0; i < n; ++i) {
if (str[i] == '0') //if the current character is 0, increase len
len++;
else
len = 0;
total_len = max(total_len, len);
}
if (total_len == n) //if the whole string has 0 only
return n * k;
int prefix = 0, suffix = 0;
for (int i = 0; str[i] == '0'; ++i, ++prefix) //find length of maximal prefix with only 0;
for (int i = n - 1; str[i] == '0'; --i, ++suffix) //find length of maximal suffix with only 0;
if (k > 1)
total_len = max(total_len, prefix + suffix);
return total_len;
}
int main() {
int k = 3;
string str = "0010010";
int res = max_length_substring(str, str.length(), k);
cout << "Maximum length of 0s: " << res;
}输出结果
Maximum length of 0s: 3