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.
分析:题目比较有意思,给两个特殊的字符,一个用0表示,另一个使用10、11表示。然后给你一个数组,要求判断这个数组最后一个字符是不是1bit。用递归法来做非常方便。
1 class Solution { 2 public boolean isOneBitCharacter(int[] bits) { 3 int cur=0; 4 return helper(bits,cur); 5 } 6 private boolean helper(int[] bits, int cur) { 7 if ( cur == bits.length-2 && bits[cur] == 1 ) return false; 8 if ( cur == bits.length - 1 ) return true; 9 return bits[cur]==0?helper(bits,cur+1):helper(bits, cur+2); 10 } 11 }