镜像数字是指一个数字倒过来还是自身，例如数字101。前20个镜像数字分别是：
0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 22, 33, 44, 55, 66, 77, 88, 99, 101。我们的任务是找出第n个镜像数字，默认所有的镜像数字默认从0开始。示例：
输入：n=1，返回：0。因为第1个镜像数字是0。
输入：n=10，返回：9。题目难度：中等
题目来源：codewars
def find_mirror_num(n: int) -> int:
pass
assert find_mirror_num(2) == 1
assert find_mirror_num(10) == 9
assert find_mirror_num(100) == 909
def find_mirror_num(n: int) -> int:
result = count = 0
while True:
if str(result)[::-1] == str(result):
count = count + 1
if count == n:
return result
result = result + 1
assert find_mirror_num(2) == 1
assert find_mirror_num(10) == 9
assert find_mirror_num(100) == 909
参考题解：
def find_mirror_num(n: int) -> int:
def gen_mirrors(x: int):
i, count = 0, 1
while True:
if count > x:
return
if str(i) == str(i)[::-1]:
yield i
count += 1
i += 1
return list(gen_mirrors(n))[-1]
assert find_mirror_num(2) == 1
assert find_mirror_num(10) == 9
assert find_mirror_num(100) == 909
思路：使用生成器generator。内部定义一个生成器函数gen_mirrors，接收一个参数x，表示需要生成多少个符合镜像条件的元素，然后转成包含n个元素的列表后，的末尾那个就是要找的第n个元素了。
比如我想找「第2个」镜像数字，那么生成器就从0开始遍历判断，一共找2个符合条件的，分别是0和1，那么第2个镜像数字就是1。
def find_mirror_num(n: int) -> int:
num_dict=dict()
num=0
while n:
if str(num)[::-1] == str(num):
num_dict[n]=num
n=n-1
num = num + 1
return num_dict[1]
public Integer findMirrorNum(Integer integer) {
int step = 0;
int start = 0;
String res = null;
StringBuilder sb;
while (step != integer) {
res = Integer.toString(start);
sb = new StringBuilder(res).reverse();
if (res.equals(sb.toString())) {
step++;
}
start++;
}
return Integer.valueOf(res);
}
@Test
public void testFindMirrorNum() {
assert findMirrorNum(2).equals(1);
assert findMirrorNum(10).equals(9);
assert findMirrorNum(100).equals(909);
}
def find_mirror_num(n: int) -> int:
m,s = 0,0
while True:
if str(m) == str(m)[::-1]:
s += 1
if s == n:
return m
m += 1
def find_mirror_num(n: int) -> int:
def get_mirror_list(x: int):
count = 1
result = 0
while count <= x:
if ''.join(reversed(str(result))) == str(result):
yield result
count += 1
result += 1
return list(get_mirror_list(n))[-1]
assert find_mirror_num(2) == 1
assert find_mirror_num(10) == 9
assert find_mirror_num(100) == 909
def find_mirror_num(n: int) -> int:
num ,count = 0,0
while (1):
nList = list(str(num))
nList.reverse()
reverseNum = ''.join(nList)
if num == int(reverseNum):
if count+1 == n:
break
else:
count += 1
num +=1
return num
def find_mirror_num(n: int) -> int:
num = 0
mirror_index = 0
while True:
for i in range(len(str(num))//2):
if str(num)[i-1] == str(num)[-i]:
...
else:
break
else:
mirror_index += 1
if mirror_index == n:
return num
num += 1
assert find_mirror_num(2) == 1
assert find_mirror_num(10) == 9
assert find_mirror_num(100) == 909
def find_mirror_num(n: int) -> int:
if n<=10:
return n-1
elif n<=19:
return (n-10)*11
else:
n = n - 19
init_ln = 3 #数字的位数
while n > (9 * 10**(math.ceil(init_ln/2)-1)):
n -= 9 * 10**(math.ceil(init_ln/2)-1)
init_ln += 1
else:
if n-1<10**(math.ceil(init_ln/2)-1):
l1=(math.ceil(init_ln/2) -len(str(n-1)))*['0']+list(str(n-1))
else:
l1 = list(str(n - 1))
l1[0] = str(int(l1[0])+1)
if init_ln%2 != 0:
l1 = l1 + list(reversed(l1[:-1]))
else:
l1 = l1 + list(reversed(l1))
return int(''.join(l1))