标签：for another turn temp ++ one eth cte equal
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.
My Solution:
class Solution { public boolean isOneBitCharacter(int[] bits) { int len = bits.length; for(int i = 0;i < len;i++){ if(i == len - 2){ if(bits[i] == 1){ return false; } if(bits[i] == 0){ return true; } } if(bits[i] == 1){ String temp = bits[i] + "" + bits[i+1]; if(temp.equals("10") || temp.equals("11")){ i++; } } } return true; } }
Another Better Solution:
class Solution { public boolean isOneBitCharacter(int[] bits) { int i = 0; while (i < bits.length - 1) { i += bits[i] + 1; } return i == bits.length - 1; } }
总结:我们注意到’0’和’10’,’11’是”互斥的”,即没有’01’这个干扰，所有当前位为’0’,则跳过一位，当前位为’1’，则跳过两位,最后通过i判断，如果i=bits.length-1，说明最后位为0,那么返回true,如果i=bits.length,说明最后可以返回二位的的字符，即为false
leetCode-1-bit and 2-bit Characters
标签：for another turn temp ++ one eth cte equal
原文地址：http://www.cnblogs.com/kevincong/p/7875288.html
