边工作边刷题：70天一遍leetcode: day 75-2
Strobogrammatic Number I/II/III
要点：记题，注意轴对称和点对称的区别。这题就是几个固定digit之间的palindrome
I
https://repl.it/CqLu
II
https://repl.it/CivO (java)
https://repl.it/CqKC (python)
- 外循环中止条件：2个或3个字符都是循环一次，每层n-2，所以是n>1
- 00不考虑的情况是n>3，因为是从内向外循环，2或者3是最外一层
- python: string可以unpack为单一char，但是变量个数必须和len(string)一样
III
https://repl.it/CkFM/2 (python iterate all results，只能beat 19.05%，懒得看快的方法了 https://discuss.leetcode.com/topic/50073/python-46ms-96-77-without-generating-all-the-numbers）
- II中的recursion就是从最短开始build到某个长度，而题的目标就是找到在low和high长度之间]的所有。所以不用外层的loop，单一的recursion就够。递归的过程就是不断增加长度，所以中途检查是不是在low/high的长度范围，同时是不是value在范围内
- positive and negative conditions:
- no positive as you keep counting until all paths fail,
- negative: two conditions:
- if len(path)+2>high.size() (every round you add two chars) or
- == but >int(high)
- 落入[low,high]之间的都有机会++res，另外need to eliminate two ‘0’ at outmost layer (unlike II, to simplify, still can recurse into it and just don’t count it)
- 中止条件：(1) >high.size() (比low小为什么也return呢？）(2) 等于high.size()但值超过
- string表示的数比较不要慌，python简单搞定int()
# A strobogrammatic number is a number that looks the same when rotated 180 degrees (looked at upside down).
# Write a function to determine if a number is strobogrammatic. The number is represented as a string.
# For example, the numbers "69", "88", and "818" are all strobogrammatic.
# Hide Company Tags Google
# Hide Tags Hash Table Math
# Hide Similar Problems (M) Strobogrammatic Number II (H) Strobogrammatic Number III
class Solution(object):
def isStrobogrammatic(self, num):
"""
:type num: str
:rtype: bool
"""
umap = {'1':'1', '8':'8', '6':'9', '9':'6', '0':'0'}
mid = ['1', '8', '0']
i,j = 0, len(num)-1
if len(num)>1 and num[0]=='0': return False
while i<=j:
if i==j:
return num[i] in mid
else:
if num[i] not in umap or umap[num[i]]!=num[j]:
return False
i+=1
j-=1
return True
import java.util.*;
class Main {
public static void main(String[] args) {
Solution sol = new Solution();
List<String> res = sol.findStrobogrammatic(5);
for(String s : res) {
System.out.println(s);
}
}
}
class Solution {
public List<String> findStrobogrammatic(int n) {
int[] nums = new int[]{1,8,6,9};
List<String> solutions = new ArrayList<>();
StringBuilder sb = new StringBuilder();
stroboHelper(nums, 0, n/2, sb, solutions);
// List<String> res = new ArrayList<>();
// for(String s : solutions) {
// }
return solutions;
}
void stroboHelper(int[] nums, int start, int n, StringBuilder sb, List<String> solutions) {
if(start==n) {
solutions.add(sb.toString());
return;
}
for(int i : nums) {
sb.append(Integer.toString(i));
stroboHelper(nums, start+1, n, sb, solutions);
sb.setLength(sb.length()-1);
}
}
}
# A strobogrammatic number is a number that looks the same when rotated 180 degrees (looked at upside down).
# Find all strobogrammatic numbers that are of length = n.
# For example,
# Given n = 2, return ["11","69","88","96"].
# Hint:
# Try to use recursion and notice that it should recurse with n - 2 instead of n - 1.
# Hide Company Tags Google
# Hide Tags Math Recursion
# Hide Similar Problems (E) Strobogrammatic Number (H) Strobogrammatic Number III
class Solution(object):
def findStrobogrammatic(self, n):
"""
:type n: int
:rtype: List[str]
"""
umap = {'1':'1', '8':'8', '6':'9', '9':'6', '0':'0'}
mid = ['1', '8', '0']
def helper(n, res, solutions):
if n<=0:
if not n and (len(res)==1 or res[0]!='0'):
solutions.append(res)
return
for k in umap.keys():
helper(n-2, k+res+umap[k], solutions)
solutions = []
if n%2==1:
for i in mid:
helper(n-1, i, solutions)
else:
helper(n, "", solutions)
return solutions
sol = Solution()
assert sol.findStrobogrammatic(2)==["11","88","96","69"]
assert sol.findStrobogrammatic(1)==["1","8","0"]
# A strobogrammatic number is a number that looks the same when rotated 180 degrees (looked at upside down).
# Write a function to count the total strobogrammatic numbers that exist in the range of low <= num <= high.
# For example,
# Given low = "50", high = "100", return 3. Because 69, 88, and 96 are three strobogrammatic numbers.
# Note:
# Because the range might be a large number, the low and high numbers are represented as string.
# Hide Tags Math Recursion
# Hide Similar Problems (E) Strobogrammatic Number (M) Strobogrammatic Number II
class Solution(object):
def strobogrammaticInRange(self, low, high):
"""
:type low: str
:type high: str
:rtype: int
"""
umap = {'1':'1', '8':'8', '6':'9', '9':'6', '0':'0'}
mid = ['', '1', '8', '0']
self.count = 0
def helper(low, high, res):
l = len(res)
if len(low) <= l <= len(high):
if l==len(high) and int(res)>int(high): return
if int(res)>=int(low) and (l==1 or res[0]!='0'):
self.count+=1
#print res
if l+2 > len(high):
return
for k in umap.keys():
helper(low, high, k+res+umap[k])
solutions = []
for m in mid:
helper(low, high, m)
return self.count
sol = Solution()
assert sol.strobogrammaticInRange("50","100")==3
assert sol.strobogrammaticInRange("12","1200")==17
