LWC 56:717. 1-bit and 2-bit Characters

LWC 56:717. 1-bit and 2-bit Characters

传送门:717. 1-bit and 2-bit Characters

Problem:

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.

思路: 如果遇到1,则可以把连带的[1,1]或者[1,0]取出来,如果遇到0,则只需要取一位,所以取到第n-2位,那么必然最后剩下一位或者零位,剩下一位返回true,剩下零位返回false.

代码如下:

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

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