leetcode-696. Count Binary Substrings

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.

题目大意：求给定字符串S中的子串，子串满足：0和1配对出现，并且所有的0或者1都是连续出现的。

解法：（这里给出的是官方的解法的翻译，还是很容易看懂的）

我们可以将字符串s转换为表示字符串中相同字符的连续块的长度的数组组。例如，如果s =“110001111000000”，则groups = [2，3，4，6]。 对于形式为'0'* k +'1'* k或'1'* k +'0'* k的每个二进制字符串，此字符串的中间必须出现在两个group之间。 我们尝试计算group[i]和group[i + 1]之间的有效二进制字符串的数量。如果我们有group[i] = 2，group[i + 1] = 3，则表示“00111”或“11000”。我们明确地可以使min（group[i]，group[i + 1]）在此字符串内有效的二进制字符串。因为这个字符串左边或右边的二进制数字必须在边界改变，所以我们的答案永远不会更大。

class Solution {
    public int countBinarySubstrings(String s) {
        int[] arr = new int[s.length()];
        int t=0;
        arr[0]=1;//保证01或者10的情况
        for(int i = 0;i

转载于:https://www.cnblogs.com/linkcode/p/8150791.html