README_EN.md

717. 1-bit and 2-bit Characters

中文文档

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 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.

Solutions

Python3

class Solution:
    def isOneBitCharacter(self, bits: List[int]) -> bool:
        i, n = 0, len(bits)
        while i < n - 1:
            i += bits[i] + 1
        return i == n - 1

Java

class Solution {
    public boolean isOneBitCharacter(int[] bits) {
        int i = 0, n = bits.length;
        while (i < n - 1) {
            i += bits[i] + 1;
        }
        return i == n - 1;
    }
}

C++

class Solution {
public:
    bool isOneBitCharacter(vector<int>& bits) {
        int i = 0, n = bits.size();
        while (i < n - 1) i += bits[i] + 1;
        return i == n - 1;
    }
};

Go

func isOneBitCharacter(bits []int) bool {
	i, n := 0, len(bits)
	for i < n-1 {
		i += bits[i] + 1
	}
	return i == n-1
}

JavaScript

/**
 * @param {number[]} bits
 * @return {boolean}
 */
var isOneBitCharacter = function (bits) {
    let i = 0;
    const n = bits.length;
    while (i < n - 1) {
        i += bits[i] + 1;
    }
    return i == n - 1;
};

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