LeetCode 696. Count Binary Substrings

Given a binary string s, return the number of non-empty 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:

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Input: s = "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:

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

Constraints:

1 <= s.length <= 105
s[i] is either '0' or '1'.

这题给了一个由0和1组成的binary string，要求string中有多少个substring有相同个数的0和1并且所有0和所有1都在一起。

拿到这道题就是一个愣住，不会。于是就直接抄了答案，于是就被这个解法惊呆了。是什么样的大天才才能想到这种解法啊！

我们计算每组连续的0或者1有多少个，相邻的两组里，长度更短的就是这两组放在一起时能拿到的最多的符合条件的substring了。就这么一想确实很合理很简单，但是自己为什么就是想不到呢。我好菜啊.jpg

写起代码来也是非常简单了，就是需要注意一下最后不能直接return result，因为最后一组还没算呢，这个小坑差点掉进去了。

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class Solution {
    public int countBinarySubstrings(String s) {
        int prev = 0;
        int curr = 1;
        int result = 0;
        for (int i = 1; i < s.length(); i++) {
            if (s.charAt(i - 1) == s.charAt(i)) {
                curr++;
            } else {
                result += Math.min(prev, curr);
                prev = curr;
                curr = 1;
            }
        }
        return result + Math.min(prev, curr);
    }
}