Leetcode - Strobogrammatic Number-优快云博客

本文链接：https://blog.youkuaiyun.com/iteye_17352/article/details/82639409

本文深入分析斯特罗博格马数字的概念，提供两种高效的算法用于判断一个数字字符串是否为斯特罗博格马数字。通过比较字符在旋转180度后的对应关系，实现快速判断。同时，文章提供了简洁的Java代码实现，便于理解和实践。

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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.
[分析]
把数字颠倒180度后看仍然一样的有0,1,8三个，另外69,96这种颠倒后也一样，需特殊处理。
因此从中间往两端或者从两端往中间逐个判断对称下标处字符，注意长度为奇数时要处理中间元素。讨论区有牛人分享了4行代码的方案，awesome！
[ref]
[url]https://leetcode.com/discuss/50594/4-lines-in-java[/url]


public class Solution {
    // Method 2
    public boolean isStrobogrammatic(String num) {
        if (num == null) return false;
        for (int i = 0, j = num.length() - 1; i <= j; i++, j--) {
            if (!"00 11 88 696".contains(num.charAt(i) + "" + num.charAt(j)))
                return false;
        }
        return true;
    }
    // Method 1
    private HashSet<Character> upDownSameDigit = new HashSet<Character>();
    public Solution() {
        upDownSameDigit.add('0');
        upDownSameDigit.add('1');
        upDownSameDigit.add('8');
    }
    public boolean isStrobogrammatic1(String num) {
        if (num == null || num.length() == 0)
            return false;
        int mid = num.length() / 2;
        if (num.length() % 2 == 1) {
            if (!upDownSameDigit.contains(num.charAt(mid)))
                return false;
            return checkSymmetrically(num, mid - 1, mid + 1);
        } else {
            return checkSymmetrically(num, mid - 1, mid);
        }
    }
    public boolean checkSymmetrically(String num, int i, int j) {
        while (i >= 0) {
            char left = num.charAt(i);
            char right = num.charAt(j);
            boolean okCase1 = left == right && upDownSameDigit.contains(left);
            boolean okCase2 = (left == '6' && right == '9') || (left == '9' && right == '6');
            if (!okCase1 && !okCase2)
                return false;
            i--; j++;
        }
        return true;
    }
}