[LeetCode] 246. Strobogrammatic Number 对称数

Posted 2020-10-26 轻风舞动

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

Python: wo

class Solution():
    def strobogrammatic(self, s):
        lookup = {\'1\': \'1\', \'8\': \'8\', \'0\': \'0\', \'6\': \'9\', \'9\': \'6\'}
        i, j = 0, len(s) - 1
        while i <= j:
            if s[i] not in lookup or  lookup[s[i]] != s[j]:
                return False
            i += 1
            j -= 1

        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 

 

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