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算法笔记: 力扣#717 1比特与2比特字符

admin 11月 12, 2020 0

问题描述

解法

分析

Python 实现

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 实现

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;
    }
}

时间复杂度

O(n).

空间复杂度

O(1).

链接

717. 1-bit and 2-bit Characters
717. 1比特与2比特字符
 (English version) Algorithm Notes: Leetcode#717 1-bit and 2-bit Characters