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难度：Easy题目：717. 1-bit and 2-bit 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: TrueExplanation: 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: FalseExplanation: 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.

翻译：给定两种字符，一种用1位的0表示，另一种用2位10或者11表示。现在给出一个由很多位字符表示的字符串，判断最后一位是不是1位字符。

思路：从头到尾遍历，如果该位数字为1，则向后前进两位，否则前进1位，循环的条件是i < n-1，即留出最后一位，所以当循环退出后，当i正好停留在n-1上，说明最后一位是单独分割开的。

参考代码：

Java

class Solution {    public boolean isOneBitCharacter(int[] bits) {        int n = bits.length;        int i = 0;        while(i