246. Strobogrammatic Number

问题描述

A strobogrammatic number is a number that looks the same when rotated 180 degrees (looked at upside down).

Write a function to determine if a number is strobogrammatic. The number is represented as a string.

For example, the numbers “69”, “88”, and “818” are all strobogrammatic.

题目链接：

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思路分析

判断一个string表示的数中心旋转180度后是否是对称的。

有这样性质的数只有0，1，6，8，9。从两头开始判断，先判断是不是这五个数，然后看相对应是否有错。

代码

递归

class Solution {
public:
    bool isStrobogrammatic(string num) {
        int n = num.length();
        for (int i = 0; i < n; i++){
            if(num[i] == '0' || num[i] == '1' ||num[i] == '6' ||num[i] == '8' ||num[i] == '9'){
            if(num[i] == '0' && num[n - 1 - i] != '0')
                return false;
            if(num[i] == '1' && num[n - 1 - i] != '1')
                return false;
            if(num[i] == '6' && num[n - 1 - i] != '9')
                return false;
            if(num[i] == '8' && num[n - 1 - i] != '8')
                return false;
            if(num[i] == '9' && num[n - 1 - i] != '6')
                return false;
            }
            else
                return false;
        }
        return true;
    }
};

时间复杂度：$O(n)$
空间复杂度：$O(1)$

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反思

挺丑陋的，一大堆if判断，看Discussion可以用hash table，建立数字之间的映射，速度更快，代码更简洁。

class Solution {
public:
    bool isStrobogrammatic(string num) {
        unordered_map<char, char> map;
        map['0'] = '0';
        map['1'] = '1';
        map['6'] = '9';
        map['8'] = '8';
        map['9'] = '6';
        int i = 0, j = num.length() - 1;
        while(i < j){
            if (map[num[i]] != num[j])
                return false;
            i++;
            j--;
        }
        if (i == j)
            return num[i] == '0'|| num[i] == '1'|| num[i] == '8';
        else
            return true;
    }
};