LeetCode | 717. 1-bit and 2-bit Characters水题

最新推荐文章于 2022-09-16 00:21:46 发布

BoomHusky

最新推荐文章于 2022-09-16 00:21:46 发布

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本文介绍了一种简单算法，用于判断由多个比特位组成的字符串是否以一比特字符结尾。通过遍历输入比特串并根据遇到的比特位决定下一步操作，最终确定最后一个字符是否为一比特长。

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canbe represented by two bits (10 or 11).

Nowgiven a string represented by several bits. Return whether the last charactermust be a one-bit character or not. The given string will always end with azero.

Example1:

Input:

bits = [1, 0, 0]

Output: True

Explanation:

The only way to decode it is two-bit character and one-bitcharacter. So the last character is one-bit character.

Example2:

Input:

bits = [1, 1, 1, 0]

Output: False

Explanation:

The only way to decode it is two-bit character and two-bitcharacter. So the last character is NOT one-bit character.

Note:

· 1 <= len(bits) <= 1000.

· bits[i] isalways 0 or 1.

水题，按照题目的要求去做就好了，遇到1进两位，遇到0进一位，最后当i==n-1说明last character一定是one-bit character

class Solution {
public:
    bool isOneBitCharacter(vector<int>& bits) {
        int n=bits.size();
        int i=0;
        while(i<n-1)
        {
            if(bits[i]==1) i+=2;
            else i+=1;
        }
        return i==n-1;
    }
};