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.  ``` var?countBinarySubstrings?=?function(s)?{ let?n?=?0,?arr?=?s.match(/([1]+)|([0]+)/g) for?(let?i?=?0;?i?<?arr.length?-?1;?i++)?{ ????????n?+=?Math.min(arr[i].length,?arr[i?+?1].length) ????} return?n }; ``` ``` // 統計前一種數連續的個數，和當前數連續的個數， // 每次數字變了結果加上這兩個里小的一個， // 同時前一個數的個數等于當前數的個數，當前數的個數重置為1。 var countBinarySubstrings = function (s) { let ans = 0; let prev = 0; let cur = 1 for (let i = 1; i < s.length; i++) { if (s[i] !== s[i - 1]) { ans += Math.min(prev, cur) prev = cur cur = 1 } else { cur++ } } return ans + Math.min(prev, cur) }; ```