246. Strobogrammatic Number

题目：

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.

链接： http://leetcode.com/problems/strobogrammatic-number/

题解：

验证一个数是否是strobogrammatic number。我们可以用验证Palindrome的方法，从头部和尾部向中间遍历。这里因为这种数的条件比较少，所以我用了一个HashMap来保存所有合理的可能性。空间复杂度应该也可以算是O(1)的

Time Complexity - O(n)， Space Complexity - O(1)。

public class Solution {
    public boolean isStrobogrammatic(String num) {
        if(num == null || num.length() == 0)
            return false;
        int lo = 0, hi = num.length() - 1;
        Map<Character, Character> map = new HashMap<>();
        map.put('0', '0');
        map.put('1', '1');
        map.put('6', '9');
        map.put('8', '8');
        map.put('9', '6');
        
        while(lo <= hi) {
            char cLo = num.charAt(lo);
            if(!map.containsKey(cLo))
                return false;
            else if(map.get(cLo) != num.charAt(hi))
                return false;
            else {
                lo++;
                hi--;
            }
        }
        
        return true;
    }
}

 

二刷：

先建立一个查找表，然后遍历字符串的时候进行查找。表很小所以可以看做O(1)。 Stefan Pochmann还有很fancy的解法，放在reference里，很漂亮。

Java：

Time Complexity - O(n)， Space Complexity - O(1)。

public class Solution {
    public boolean isStrobogrammatic(String num) {
        if (num == null || num.length() == 0) {
            return false;
        }
        Map<Character, Character> map = new HashMap();
        map.put('6', '9');
        map.put('9', '6');
        map.put('1', '1');
        map.put('8', '8');
        map.put('0', '0');
        int lo = 0, hi = num.length() - 1;
        while (lo <= hi) {
            if (map.containsKey(num.charAt(hi)) && num.charAt(lo) == map.get(num.charAt(hi))) {
                lo++;
                hi--;
            } else {
                return false;
            }
        }
        return true;
    }
}

 

三刷：

Java:

public class Solution {
    public boolean isStrobogrammatic(String num) {
        if (num == null) return false;
        Map<Character, Character> map = new HashMap<>();
        map.put('6', '9');
        map.put('9', '6');
        map.put('8', '8');
        map.put('1', '1');
        map.put('0', '0');
        int lo = 0, hi = num.length() - 1;
        while (lo <= hi) {
            char loChar = num.charAt(lo);
            char hiChar = num.charAt(hi);
            if (!map.containsKey(loChar) || map.get(loChar) != hiChar) return false;
            lo++;
            hi--;
        }
        return true;
    }
}

 

 

 

 

Reference:

https://leetcode.com/discuss/50594/4-lines-in-java