717. 1-bit and 2-bit Characters

最新推荐文章于 2019-10-12 19:32:37 发布

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本文介绍了一种用于识别字符串中最后一个字符是否为一比特字符的算法。通过递归方式从字符串末尾开始解析，判断其构成模式。适用于由一比特（0）和两比特（10或11）字符组成的字符串。

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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.

解题思路：题目的大意是，数组由one bit 0和tow bit 10或11组成，判断最后一位是由one bit组成的，返回true，tow bit组成的返回false

两种情况：

1.首先由后往前推，如果后两位都为0的话，那肯定是true

2. 如果倒数第二位为1的话，那就得判断后面的最后一位是10产生的还是0产生的。

思路为：我们知道两位的时候是10或者11，在数组中如果碰到1，则必为two bit .则跳过1后面的那位，直接到1后面的后面那位再判定。这样一直递归下去，看最后一位是否是被跳过的。如果是被跳过，那就是two bit产生的，返回false

class Solution {
public:
	bool isOneBitCharacter(vector<int>& bits) {
		if (bits[bits.size()-1] == 0 && bits[bits.size() - 2] == 0) return true;	
		int i = 0;
		return dfs(bits, i);
	}
private:
	bool dfs(vector<int>& bits, int idx) {
		if (idx == bits.size()) return false;
		if (idx == bits.size() - 1 && bits[idx] == 0) return true;
		if (bits[idx] == 1) return dfs(bits, idx + 2);
		else
			return dfs(bits, ++idx);
	}
};