[Leetcode] 696. Count Binary Substrings 解题报告

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本文介绍了一种算法，用于统计给定二进制字符串中满足特定条件的子串数量。这些子串要求包含相同数量且连续的0和1，并且所有0和1必须连续出现。文章提供了详细的思路解析及C++实现代码。

题目：

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.

思路：

首先统计出字符串中连续出现的0的个数以及连续出现的1的个数，然后查看相邻的0和1最多可以构成多少个符合条件的子串即可。

代码：

class Solution {
public:
    int countBinarySubstrings(string s) {
        if (s.length() == 0) {
            return 0;
        }
        vector<int> counts;
        counts.push_back(1);
        for (int i = 1; i < s.length(); ++i) {
            if (s[i] == s[i - 1]) {
                ++counts.back();
            }
            else {
                counts.push_back(1);
            }
        }
        int ret = 0;
        for (int i = 0; i + 1 < counts.size(); ++i) {
            ret += min(counts[i], counts[i + 1]);
        }
        return ret;
    }
};