# Strobogrammatic 数生成

## [Strobogrammatic Number II](https://leetcode.com/problems/strobogrammatic-number-ii/)

为什么一个这么简单的 DFS 能超过 89% ..

### 注意：index == 0 并且 i == 0 的时候要跳过，免得在起始位置填上 0 .

```java
public class Solution {
    public List<String> findStrobogrammatic(int n) {
        List<String> list = new ArrayList<>();
        char[] num1 = {'0','1','8','6','9'};
        char[] num2 = {'0','1','8','9','6'};
        char[] number = new char[n];

        dfs(list, number, num1, num2, 0);

        return list;
    }

    private void dfs(List<String> list, char[] number, char[] num1, char[] num2, int index){
        int left = index;
        int right = number.length - index - 1;

        if(left > right){
            list.add(new String(number));
            return;
        }
        // We can fill in 0,1,8 only
        if(left == right){
            for(int i = 0; i < 3; i++){
                number[left] = num1[i];
                dfs(list, number, num1, num2, index + 1);
            }
        } else {
            for(int i = 0; i < num1.length; i++){
                if(index == 0 && i == 0) continue;
                number[left] = num1[i];
                number[right] = num2[i];
                dfs(list, number, num1, num2, index + 1);
            }
        }
    }
}
```

## [Strobogrammatic Number III](https://leetcode.com/problems/strobogrammatic-number-iii/)

Google 面经里的 follow-up 是，给定一个上限 n ，输出所有上限范围内的数。

办法土了点，遍历所有 lowLen \~ highLen 区间的长度，生成所有可能的结果，考虑到区间可能是大数，我们就改一下，自己写一个 String compare 函数好了。

### 后来发现有点多余，可以直接用内置的 str1.compareTo(str2).

超过 81.92% \~

```java
public class Solution {
    int count = 0;
    public int strobogrammaticInRange(String low, String high) {
        int lowLen = low.length();
        int highLen = high.length();

        char[] num1 = {'0','1','8','6','9'};
        char[] num2 = {'0','1','8','9','6'};

        for(int i = lowLen; i <= highLen; i++){
            char[] number = new char[i];
            dfs(number, num1, num2, 0, low, high);
        }

        return count;
    }

    private void dfs(char[] number, char[] num1, char[] num2, int index, String low, String high){
        int left = index;
        int right = number.length - index - 1;

        if(left > right){
            String num = new String(number);
            if(compare(low, num) <= 0 && compare(num, high) <= 0) count++;
            return;
        } else if(left == right){
            for(int i = 0; i < 3; i++){
                number[left] = num1[i];
                dfs(number, num1, num2, index + 1, low, high);
            }
        } else {
            for(int i = 0; i < 5; i++){
                if(index == 0 && i == 0) continue;
                number[left] = num1[i];
                number[right] = num2[i];
                dfs(number, num1, num2, index + 1, low, high);
            }
        }
    }

    // -1 : str1 is bigger
    // 1 : str 2 is bigger
    // 0 : equal
    private int compare(String str1, String str2){
        if(str1.length() > str2.length()) return 1;
        else if(str1.length() < str2.length()) return -1;
        else {
            for(int i = 0; i < str1.length(); i++){
                int digit1 = str1.charAt(i) - '0';
                int digit2 = str2.charAt(i) - '0';

                if(digit1 != digit2) return (digit1 > digit2) ? 1: -1;
            }
        }
        // Equal
        return 0;
    }

}
```


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