[LeetCode] 246. Strobogrammatic Number 对称数

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.

翻转180度后对称的数有：8->8, 0->0, 1->1, 6->9, 9->6，从两边向中间检查对应位置的两个数是否满足对称数就行了。比如619，先判断6和9是有映射的，然后1和自己又是映射，所以是对称数。有点像判断回文Palindrome，回文判断是否相等，这里判断是否满足那几个数字的条件。判断是可以直接写条件判断，也可以用HashMap存放数字的映射，然后用双指针从两边向中间查看。

Java:

public class Solution {
    public boolean isStrobogrammatic(String num) {
        HashMap<Character, Character> map = new HashMap<Character, Character>();
        map.put(‘1‘,‘1‘);
        map.put(‘0‘,‘0‘);
        map.put(‘6‘,‘9‘);
        map.put(‘9‘,‘6‘);
        map.put(‘8‘,‘8‘);
        int left = 0, right = num.length() - 1;
        while(left <= right){
            if(!map.containsKey(num.charAt(right)) || num.charAt(left) != map.get(num.charAt(right))){
                return false;
            }
            left++;
            right--;
        }
        return true;
    }
}

Python:

class Solution:
    lookup = {‘0‘:‘0‘, ‘1‘:‘1‘, ‘6‘:‘9‘, ‘8‘:‘8‘, ‘9‘:‘6‘}

    def isStrobogrammatic(self, num):
        n = len(num)
        for i in xrange((n+1) / 2):
            if num[n-1-i] not in self.lookup or                num[i] != self.lookup[num[n-1-i]]:
                return False
        return True

C++:

class Solution {
public:
    bool isStrobogrammatic(string num) {
        unordered_map<char, char> m {{‘0‘, ‘0‘}, {‘1‘, ‘1‘}, {‘8‘, ‘8‘}, {‘6‘, ‘9‘}, {‘9‘, ‘6‘}};
        for (int i = 0; i <= num.size() / 2; ++i) {
            if (m[num[i]] != num[num.size() - i - 1]) return false;
        }
        return true;
    }
};

类似题目：　

[LeetCode] 247. Strobogrammatic Number II 对称数II

[LeetCode] 248. Strobogrammatic Number III 对称数III

原文地址：https://www.cnblogs.com/lightwindy/p/8491309.html