We have two special characters. The first character can be represented by one bit 0. The second character can be represented by two bits (10 or 11).
Now given a string represented by several bits. Return whether the last character must be a one-bit character or not. The given string will always end with a zero.
Example 1:
Input: bits = [1, 0, 0] Output: True Explanation: The only way to decode it is two-bit character and one-bit character. So the last character is one-bit character.
Example 2:
Input: bits = [1, 1, 1, 0] Output: False Explanation: The only way to decode it is two-bit character and two-bit character. So the last character is NOT one-bit character.
Note:
1 <= len(bits) <= 1000.bits[i] is always 0 or 1.我们有两个特殊字符。第一个字符可以用1位表示。第二个字符可以用2位（10或11）表示。
现在给出一个由几位表示的字符串。返回最后一个字符是否必须是一位字符。给定的字符串将始终以零结束。
/*** @param {number[]} bits* @return {boolean}*/var isOneBitCharacter = function(bits) {let str = bits.join("").replace(/(1.)/g,"#");return str[str.length-1] == 0;};