https://leetcode-cn.com/problems/lexicographical-numbers/
示例和提示 代码:(递归)class Solution {
public:
void dfs(int i,int n,vector<int>& ret)
{
if(i>n) return;
ret.push_back(i);
for(int j = 0;j<=9;j++){
int next = i*10+j;
if(next>n) break;
dfs(next,n,ret);
}
}
vector<int> lexicalOrder(int n) {
vector<int> ret;
for(int i = 1;i<=9;i++){
dfs(i,n,ret);
}
return ret;
}
};
递归图解:
代码:迭代
class Solution {
public:
vector<int> lexicalOrder(int n) {
vector<int> ret(n);
int num = 1;
for (int i = 0; i < n; i++)
{
ret[i] = num;
if (num * 10 <= n)
num *= 10;
else
{
while (num % 10 == 9 || num + 1 > n)
{
num /= 10;
}
num++;
}
}
return ret;
}
};
迭代文字解析:
先找一个结果示例,方便对照文字
250
[1, 10, 100, 101, 102, 103, 104, 105, 106, 107, 108, 109, 11, 110,
111, 112, 113, 114, 115, 116, 117, 118, 119, 12, 120,
121, 122, 123, 124, 125, 126, 127, 128, 129, 13, 130,
131, 132, 133, 134, 135, 136, 137, 138, 139, 14, 140,
141, 142, 143, 144, 145, 146, 147, 148, 149, 15, 150,
151, 152, 153, 154, 155, 156, 157, 158, 159, 16, 160,
161, 162, 163, 164, 165, 166, 167, 168, 169, 17, 170,
171, 172, 173, 174, 175, 176, 177, 178, 179, 18, 180,
181, 182, 183, 184, 185, 186, 187, 188, 189, 19, 190,
191, 192, 193, 194, 195, 196, 197, 198, 199, 2, 20, 200,
201, 202, 203, 204, 205, 206, 207, 208, 209, 21, 210, 211,
212, 213, 214, 215, 216, 217, 218, 219, 22, 220, 221, 222,
223, 224, 225, 226, 227, 228, 229, 23, 230, 231, 232, 233,
234, 235, 236, 237, 238, 239, 24, 240,
241, 242, 243, 244, 245, 246, 247, 248, 249, 25, 250,
26, 27, 28, 29, 3, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 4, 40,
41, 42, 43, 44, 45, 46, 47, 48, 49, 5, 50,
51, 52, 53, 54, 55, 56, 57, 58, 59, 6, 60,
61, 62, 63, 64, 65, 66, 67, 68, 69, 7, 70,
71, 72, 73, 74, 75, 76, 77, 78, 79, 8, 80,
81, 82, 83, 84, 85, 86, 87, 88, 89, 9, 90,
91, 92, 93, 94, 95, 96, 97, 98, 99]
先逐步深入至最低端
1 10 100
100之后的1000不能进入
此时就开始插入
101 102 103…
当需要插入的数字等于 109或者 n时,
就需要停止插入
然后后退一步至:2
然后重复第一个步骤 20
200
2000不符合所以是201
依次循环下去
(感觉和递归有异曲同工之妙)
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