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algorithm notes: leetcode#717 1

admin 11月 12, 2020 0

Problem

Solution

Analysis

Python implementation

class :
    def isOneBitCharacter(self, bits):
        """
        :type bits: List[int]
        :rtype: bool
        """
        pointer = 0
        while pointer < len(bits) - 1:
            pointer += (2 if bits[pointer] else 1)
        return pointer == len(bits) - 1

Java implementation

class  {
    public boolean isOneBitCharacter(int[] bits) {
        int pointer = 0;
        while(pointer < bits.length - 1)
            pointer += (bits[pointer] == 1 ? 2 : 1);
        return pointer == bits.length - 1;
    }
}

Time complexity

O(n).

Space complexity

O(1).

Link

717. 1-bit and 2-bit Characters
(中文版) 算法笔记: 力扣#717 1比特与2比特字符