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717. 1-bit and 2-bit Characters

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).

Given a binary array bits that ends with 0, return true if the last character must be a one-bit character.

 

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.

 

Constraints:

• 1 <= bits.length <= 1000
• bits[i] is either 0 or 1.

Solutions

• Java
• C++
• Python
• Go
• Javascript
• TypeScript
• RenderScript

• 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;
      }
  }
  

• 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;
      }
  };
  

• 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
  
  

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

• /**
   * @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;
  };
  
  

• function isOneBitCharacter(bits: number[]): boolean {
      let i = 0;
      const n = bits.length;
      while (i < n - 1) {
          i += bits[i] + 1;
      }
      return i === n - 1;
  }
  
  

• impl Solution {
      pub fn is_one_bit_character(bits: Vec<i32>) -> bool {
          let mut i = 0usize;
          let n = bits.len();
          while i < n - 1 {
              i += (bits[i] + 1) as usize;
          }
          i == n - 1
      }
  }
  
  

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