leetcode 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.

解法1:

使用两个计数器，记录连续1和0的次数。每次循环时候，如果当前数字的计数器比之前数字的计数器数值小，则ans+=1。

计数细节：看到计数连续出现的字符，当s[i]==s[i-1]，计数器+=1，否则reset计数器，将上次的计数存起来。

class Solution(object):
    def countBinarySubstrings(self, s):
        """
        :type s: str
        :rtype: int
        """
        # 00     11         00   11
        #  |      |          |    |
        #cnt=2   new_cnt=2 cnt=2 new_cnt=2
        #    ans +=2    ans+=2   ans+=2
        ans = 0
        cnt = 1
        pre_cnt = 0
        for i in xrange(1, len(s)):
            if s[i]==s[i-1]:
                cnt += 1
            else:
                pre_cnt = cnt
                cnt = 1
            if pre_cnt >= cnt:
                ans += 1
        return ans        

直接将连续的字符串计数器用数组记录起来，然后将结果加起来！此法更直观！！

• '00001111' => [4, 4] => min(4, 4) => 4
• '00110' => [2, 2, 1] => min(2, 2) + min(2, 1) => 3
• '10101' => [1, 1, 1, 1, 1] => 4

class Solution(object):
    def countBinarySubstrings(self, s):
        cnt = 1
        chunks = []
        for i in xrange(1, len(x)):
            if s[i] == s[i-1]:
                cnt += 1
            else:
                chunks.append(cnt)
                cnt = 1
        return sum(min(chunks[i], chunks[i-1]) for i in xrange(1, len(chunks)))            

 上述解法有错，漏掉了s[-1]计数，修正后的：

class Solution(object):
    def countBinarySubstrings(self, s):
        """
        :type s: str
        :rtype: int
        """
        cnt = 1
        arr = []
        for i in range(1, len(s)):
            if s[i] == s[i-1]:
                cnt += 1
            else:
                arr.append(cnt)
                cnt = 1
        arr.append(cnt)
        return sum(min(arr[i], arr[i-1]) for i in range(1, len(arr)))