leetcode 717. 1-bit and 2-bit Characters -easy

https://leetcode.com/problems/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).

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.

Solution 1：

读清楚题目。

• 明白题目意图，就会发现，题目的意思是要判断最后一个0元素是属于0还是输入10；
• 遍历数组，给定指针，若当前位为1则指针+2；若当前位为0，则指针+1；
• 判断最后指针是否与bits.length-1相等，相等则为真，否则为假；其中length=1的情况也包括进去了。

参考：https://blog.csdn.net/koala_tree/article/details/78472100

class Solution {
    public boolean isOneBitCharacter(int[] bits) {
        int i = 0;
        while (i < bits.length-1){
            if (bits[i] == 1){
                i += 2;
            }else{
                i++;
            }
        }
        return i == bits.length-1;
    }
}

 

• 时间复杂度：O(n)，空间复杂度：O(1)

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