Skip to content

Latest commit

 

History

History
177 lines (139 loc) · 4.62 KB

File metadata and controls

177 lines (139 loc) · 4.62 KB
comments difficulty edit_url tags
true
Medium
Recursion
Array
String

中文文档

Description

Given an integer n, return all the strobogrammatic numbers that are of length n. You may return the answer in any order.

A strobogrammatic number is a number that looks the same when rotated 180 degrees (looked at upside down).

 

Example 1:

Input: n = 2
Output: ["11","69","88","96"]

Example 2:

Input: n = 1
Output: ["0","1","8"]

 

Constraints:

  • 1 <= n <= 14

Solutions

Solution 1: Recursion

If the length is $1$, then the strobogrammatic numbers are only $0, 1, 8$; if the length is $2$, then the strobogrammatic numbers are only $11, 69, 88, 96$.

We design a recursive function $dfs(u)$, which returns the strobogrammatic numbers of length $u$. The answer is $dfs(n)$.

If $u$ is $0$, return a list containing an empty string, i.e., [""]; if $u$ is $1$, return the list ["0", "1", "8"].

If $u$ is greater than $1$, we traverse all the strobogrammatic numbers of length $u - 2$. For each strobogrammatic number $v$, we add $1, 8, 6, 9$ to both sides of it, and we can get the strobogrammatic numbers of length u.

Note that if $u \neq n$, we can also add $0$ to both sides of the strobogrammatic number.

Finally, we return all the strobogrammatic numbers of length $n$.

The time complexity is $O(2^{n+2})$.

Similar problems:

Python3

class Solution:
    def findStrobogrammatic(self, n: int) -> List[str]:
        def dfs(u):
            if u == 0:
                return ['']
            if u == 1:
                return ['0', '1', '8']
            ans = []
            for v in dfs(u - 2):
                for l, r in ('11', '88', '69', '96'):
                    ans.append(l + v + r)
                if u != n:
                    ans.append('0' + v + '0')
            return ans

        return dfs(n)

Java

class Solution {
    private static final int[][] PAIRS = {{1, 1}, {8, 8}, {6, 9}, {9, 6}};
    private int n;

    public List<String> findStrobogrammatic(int n) {
        this.n = n;
        return dfs(n);
    }

    private List<String> dfs(int u) {
        if (u == 0) {
            return Collections.singletonList("");
        }
        if (u == 1) {
            return Arrays.asList("0", "1", "8");
        }
        List<String> ans = new ArrayList<>();
        for (String v : dfs(u - 2)) {
            for (var p : PAIRS) {
                ans.add(p[0] + v + p[1]);
            }
            if (u != n) {
                ans.add(0 + v + 0);
            }
        }
        return ans;
    }
}

C++

class Solution {
public:
    const vector<pair<char, char>> pairs = {{'1', '1'}, {'8', '8'}, {'6', '9'}, {'9', '6'}};

    vector<string> findStrobogrammatic(int n) {
        function<vector<string>(int)> dfs = [&](int u) {
            if (u == 0) return vector<string>{""};
            if (u == 1) return vector<string>{"0", "1", "8"};
            vector<string> ans;
            for (auto& v : dfs(u - 2)) {
                for (auto& [l, r] : pairs) ans.push_back(l + v + r);
                if (u != n) ans.push_back('0' + v + '0');
            }
            return ans;
        };
        return dfs(n);
    }
};

Go

func findStrobogrammatic(n int) []string {
	var dfs func(int) []string
	dfs = func(u int) []string {
		if u == 0 {
			return []string{""}
		}
		if u == 1 {
			return []string{"0", "1", "8"}
		}
		var ans []string
		pairs := [][]string{{"1", "1"}, {"8", "8"}, {"6", "9"}, {"9", "6"}}
		for _, v := range dfs(u - 2) {
			for _, p := range pairs {
				ans = append(ans, p[0]+v+p[1])
			}
			if u != n {
				ans = append(ans, "0"+v+"0")
			}
		}
		return ans
	}
	return dfs(n)
}