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.struct Solution;
impl Solution {
fn is_one_bit_character(bits: Vec<i32>) -> bool {
let mut i = 0;
let n = bits.len();
let mut one_bit: Option<bool> = None;
while i < n {
if bits[i] == 1 {
i += 2;
one_bit = Some(false);
} else {
i += 1;
one_bit = Some(true);
}
}
if let Some(res) = one_bit {
res
} else {
false
}
}
}
#[test]
fn test() {
let bits = vec![1, 0, 0];
assert_eq!(Solution::is_one_bit_character(bits), true);
let bits = vec![1, 1, 0, 0];
assert_eq!(Solution::is_one_bit_character(bits), true);
}