696. Count Binary Substrings 计数二进制子字符串

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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.

    • 给一个字符串s，计算具有相同数字0和1的非空（连续）子字符串的数量，并且这些子字符串中的所有0和所有1都被连续分组。 发生多次的子字符串将被计数它们发生的次数。

    1. /**
    2.  * @param {string} s
    3.  * @return {number}
    4.  */
    5. var countBinarySubstrings = function (s) {
    6.     let prevRunLength = 0;
    7.     let curRunLength = 1;
    8.     let res = 0;
    9.     for (let i = 1; i < s.length; i++) {
    10.         if (s[i] == s[i - 1]) {
    11.             curRunLength++;
    12.         } else {
    13.             prevRunLength = curRunLength;
    14.             curRunLength = 1;
    15.         }
    16.         if (prevRunLength >= curRunLength) res++;
    17.     }
    18.     return res;
    19. };
    21. //console.log(countBinarySubstrings("00110011"));





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