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Description
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
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.
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.
Solution
当前数字如果是1的话，则下一个数字可以为0或者1，将下标往后移两位；如果当前数字是0的话，则判断当前位置是不是最后一位，如果是，则答案题目要求，返回true，如果不是，则继续将位置向后移动一位。
class Solution {
public boolean isOneBitCharacter(int[] bits) {
int i = 0;
while(i < bits.length) {
if(bits[i] == 1) {
i+=2;
} else {
if(i == bits.length - 1)
return true;
i++;
}
}
return false;
}
}