https://leetcode.com/problems/1-bit-and-2-bit-characters/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 always0
or1
.
Solution 1:
读清楚题目。
- 明白题目意图,就会发现,题目的意思是要判断最后一个
0
元素是属于0
还是输入10
; - 遍历数组,给定指针,若当前位为
1
则指针+2
;若当前位为0
,则指针+1
; - 判断最后指针是否与
bits.length-1
相等,相等则为真,否则为假;其中length=1的情况也包括进去了。
参考:https://blog.csdn.net/koala_tree/article/details/78472100
class Solution {
public boolean isOneBitCharacter(int[] bits) {
int i = ;
while (i < bits.length-){
if (bits[i] == ){
i += ;
}else{
i++;
}
}
return i == bits.length-;
}
}
- 时间复杂度:O(n),空间复杂度:O(1)