[LeetCode] 1-bit and 2-bit Characters

标签：int   public   vector   数字   whether   false   bit   判断   lag   

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.

我们有两个特殊字符。第一个字符可以由一个位0表示。第二个字符可以由两个位（10或11）表示。

现在给出一个由几位表示的字符串。返回最后一个字符是否必须是一位字符。给定的字符串将始终以零结尾。

思路：使用蛮力判断每一个特殊字符的，如果是两个10、11，则第一位一定是1，让bit数组跳过一个字符即i=i+2

如果是一个字符0，则让bit数组继续遍历0后的下一个字符。直到判断到倒数第二个数字为止，如果这个数字是1，则返回false，如果这个数字是0，则返回true。

class Solution {
public:
    bool isOneBitCharacter(vector<int>& bits) {
        int n = bits.size(), flag = true;
        for (int i = 0; i != n;) {
            if (bits[i] == 1) {
                if (i == n - 2) {
                    flag = false;
                    break;
                }
                else
                    i = i + 2;
            }
            else {
                if (i == n - 2) {
                    flag = true;
                    break;
                }
                else
                    i = i + 1;
            }
        }
        return flag;
    }
};
// 3 ms

[LeetCode] 1-bit and 2-bit Characters

标签：int   public   vector   数字   whether   false   bit   判断   lag   

原文地址：http://www.cnblogs.com/immjc/p/7761904.html