696. Count Binary Substrings

Give a string `s`, count the number of non-empty (contiguous) substrings that have the same number of 0’s and 1’s, and all the 0’s and all the 1’s in these substrings are grouped consecutively.

Substrings that occur multiple times are counted the number of times they occur.

Example 1:

``````Input: "00110011"
Output: 6
Explanation: There are 6 substrings that have equal number of consecutive 1's and 0's: "0011", "01", "1100", "10", "0011", and "01".

Notice that some of these substrings repeat and are counted the number of times they occur.

Also, "00110011" is not a valid substring because all the 0's (and 1's) are not grouped together.
``````

Example 2:

``````Input: "10101"
Output: 4
Explanation: There are 4 substrings: "10", "01", "10", "01" that have equal number of consecutive 1's and 0's.
``````

Note:

• `s.length` will be between 1 and 50,000.
• `s` will only consist of “0” or “1” characters.

``````class Solution {
public:
int countBinarySubstrings(string s) {
int zeros = 0, ones = 0, res = 0;
for (int i = 0; i < s.size(); ++i) {
if (i == 0) {
(s[i] == '1') ? ++ones : ++zeros;
} else {
if (s[i] == '1') {
ones = (s[i - 1] == '1') ? ones + 1 : 1;
if (zeros >= ones) ++res;
} else if (s[i] == '0') {
zeros = (s[i - 1] == '0') ? zeros + 1 : 1;
if (ones >= zeros) ++res;
}
}
}
return res;
}
};
``````

``````class Solution {
public:
int countBinarySubstrings(string s) {
int res = 0, pre = 0, cur = 1, n = s.size();
for (int i = 1; i < n; ++i) {
if (s[i] == s[i - 1]) ++cur;
else {
pre = cur;
cur = 1;
}
if (pre >= cur) ++res;
}
return res;
}
};
``````

Encode and Decode Strings

https://discuss.leetcode.com/topic/107096/java-o-n-time-o-1-space

LeetCode All in One 题目讲解汇总(持续更新中…)

 微信打赏 Venmo 打赏
（欢迎加入博主的知识星球，博主将及时答疑解惑，并分享刷题经验与总结，试运营期间前五十位可享受半价优惠～）

×

Help us with donation