1513. Number of Substrings With Only 1s

Given a binary string s (a string consisting only of '0' and '1's).

Return the number of substrings with all characters 1's.

Since the answer may be too large, return it modulo 10^9 + 7.

Example 1:

Input: s = "0110111"
Output: 9
Explanation: There are 9 substring in total with only 1's characters.
"1" -> 5 times.
"11" -> 3 times.
"111" -> 1 time.

Example 2:

Input: s = "101"
Output: 2
Explanation: Substring "1" is shown 2 times in s.

Example 3:

Input: s = "111111"
Output: 21
Explanation: Each substring contains only 1's characters.

Example 4:

Input: s = "000"
Output: 0

Constraints:

• s[i] == '0' or s[i] == '1'
• 1 <= s.length <= 10^5]

class Solution {
    public int numSub(String s) {
        int mod = 1000000007;
        int len = 0;
        int res = 0;
        char[] arr = s.toCharArray();
        for(int i = 0; i < arr.length; i++){
            if(arr[i] == '1'){
                len++;
                res = (res + len) % mod;
            }
            else len = 0;
        }
        return res % mod;
    }
}

看连续1的数量，比如111，res就是1+2+3，遇到0就重置0

总结：

We see that for a string with all 1 like 111, it has substring of all 1's like 1, 1,1 11, 11, 111. Totally 6 which is made by 1 + 2 + 3

So we iterate the string, when it's 1 we make len++, res = (res+len)%mod, when it's 0 we make len = 0. Finally we get the answer.