[LeetCode] 696. Count Binary Substrings_Easy

利用group, 将每个连着的0或者1计数并且append进入group里面, 然后再将group里面的两两比较, 得到min, 并且加入到ans即可.   T: O(n)   S: O(n)  比较好理解

improve: 思路相同, 用pre和cur去将space节省到O(1)

Code

1)  T: O(n)   S: O(n)

class Solution:
    def countBinarySubstrings(self, s):
        group = [1]
        for i in range(1,len(s)):
            if s[i] != s[i-1]:
                group.append(1)
            else:
                group[-1] += 1
        ans = 0
        for i in range(len(group) -1):
            ans += min(group[i], group[i+1])
        return ans

2) T: O(n)   S: O(1)

class Solution:
    def countBinarySubstrings(self, s):
        ans, pre, cur = 0, 0, 1
        for i in range(1,len(s)):
            if s[i] != s[i-1]:
                ans += min(pre, cur)
                pre, cur = cur, 1
            else:
                cur += 1
        return ans + min(pre, cur)