• 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
    
    
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  • 原文地址:https://www.cnblogs.com/absolute/p/5815684.html
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