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386. 字典序排数

题目描述

给你一个整数 n ，按字典序返回范围 [1, n] 内所有整数。

你必须设计一个时间复杂度为 O(n) 且使用 O(1) 额外空间的算法。

 

示例 1：

输入：n = 13
输出：[1,10,11,12,13,2,3,4,5,6,7,8,9]

示例 2：

输入：n = 2
输出：[1,2]

 

提示：

  • 1 <= n <= 5 * 104

解法

方法一：DFS

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class Solution:
    def lexicalOrder(self, n: int) -> List[int]:
        def dfs(u):
            if u > n:
                return
            ans.append(u)
            for i in range(10):
                dfs(u * 10 + i)

        ans = []
        for i in range(1, 10):
            dfs(i)
        return ans

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class Solution {
    public List<Integer> lexicalOrder(int n) {
        List<Integer> ans = new ArrayList<>();
        for (int i = 1; i < 10; ++i) {
            dfs(i, n, ans);
        }
        return ans;
    }

    private void dfs(int u, int n, List<Integer> ans) {
        if (u > n) {
            return;
        }
        ans.add(u);
        for (int i = 0; i < 10; ++i) {
            dfs(u * 10 + i, n, ans);
        }
    }
}

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class Solution {
public:
    vector<int> lexicalOrder(int n) {
        vector<int> ans;
        for (int i = 1; i < 10; ++i) dfs(i, n, ans);
        return ans;
    }

    void dfs(int u, int n, vector<int>& ans) {
        if (u > n) return;
        ans.push_back(u);
        for (int i = 0; i < 10; ++i) dfs(u * 10 + i, n, ans);
    }
};

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func lexicalOrder(n int) []int {
    var ans []int
    var dfs func(u int)
    dfs = func(u int) {
        if u > n {
            return
        }
        ans = append(ans, u)
        for i := 0; i < 10; i++ {
            dfs(u*10 + i)
        }
    }
    for i := 1; i < 10; i++ {
        dfs(i)
    }
    return ans
}

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impl Solution {
    fn dfs(mut num: i32, n: i32, res: &mut Vec<i32>) {
        if num > n {
            return;
        }
        res.push(num);
        for i in 0..10 {
            Self::dfs(num * 10 + i, n, res);
        }
    }

    pub fn lexical_order(n: i32) -> Vec<i32> {
        let mut res = vec![];
        for i in 1..10 {
            Self::dfs(i, n, &mut res);
        }
        res
    }
}

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/**
 * @param {number} n
 * @return {number[]}
 */
var lexicalOrder = function (n) {
    let ans = [];
    function dfs(u) {
        if (u > n) {
            return;
        }
        ans.push(u);
        for (let i = 0; i < 10; ++i) {
            dfs(u * 10 + i);
        }
    }
    for (let i = 1; i < 10; ++i) {
        dfs(i);
    }
    return ans;
};

方法二

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class Solution {
    public List<Integer> lexicalOrder(int n) {
        List<Integer> ans = new ArrayList<>();
        int v = 1;
        for (int i = 0; i < n; ++i) {
            ans.add(v);
            if (v * 10 <= n) {
                v *= 10;
            } else {
                while (v % 10 == 9 || v + 1 > n) {
                    v /= 10;
                }
                ++v;
            }
        }
        return ans;
    }
}

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class Solution {
public:
    vector<int> lexicalOrder(int n) {
        vector<int> ans;
        int v = 1;
        for (int i = 0; i < n; ++i) {
            ans.push_back(v);
            if (v * 10 <= n)
                v *= 10;
            else {
                while (v % 10 == 9 || v + 1 > n) v /= 10;
                ++v;
            }
        }
        return ans;
    }
};

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func lexicalOrder(n int) []int {
    var ans []int
    v := 1
    for i := 0; i < n; i++ {
        ans = append(ans, v)
        if v*10 <= n {
            v *= 10
        } else {
            for v%10 == 9 || v+1 > n {
                v /= 10
            }
            v++
        }
    }
    return ans
}

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