摘要:题目解答题目解答先考虑最底层的两种情况,当和当的时候,就是最中间的数为空还是存在唯一的一个数。然后我们在这个基础上,用循环两个数两个数地一起向外扩张。扩张后的结果存在里,作为再服务于上一层的扩张,得到最终结果。
246.Strobogrammatic NumberI
题目:
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.
解答:
public class Solution { //1, 6, 8, 9, 0 public boolean isStrobogrammatic(String num) { Mapmap = new HashMap<>(); map.put(1, 1); map.put(6, 9); map.put(8, 8); map.put(9, 6); map.put(0, 0); int len = num.length(); for (int i = 0; i < len; i++) { int c1 = num.charAt(i) - "0", c2 = num.charAt(len - 1 - i) - "0"; if (!map.containsKey(c1) || !map.containsKey(c2) || c1 != map.get(c2)) { return false; } } return true; } }
247.StroboGrammatic NumberII
题目:
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"].
解答:
先考虑最底层的两种情况,当n == 0和当n == 1的时候,就是最中间的数为空还是存在唯一的一个数。然后我们在这个基础上,用循环两个数两个数地一起向外扩张。扩张后的结果存在result里,作为base再服务于上一层的扩张,得到最终结果。
public class Solution { public Listhelper(int n, int m) { if (n == 0) return new ArrayList (Arrays.asList("")); if (n == 1) return new ArrayList (Arrays.asList("0", "1", "8")); List list = helper(n - 2, m); List result = new ArrayList (); for (int i = 0; i < list.size(); i++) { String s = list.get(i); if (n != m) result.add("0" + s + "0"); result.add("1" + s + "1"); result.add("8" + s + "8"); result.add("6" + s + "9"); result.add("9" + s + "6"); } return result; } public List findStrobogrammatic(int n) { return helper(n, n); } }
248.Strobogrammatic NumberIII
题目:
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.
解答:
有了上一题作基础,这里我们可以先求出所有长度满足的数,再通过与low,high的compare比较,选出最终的结果并count。
public class Solution { public Listhelper(int n, int max) { if (n == 0) return new ArrayList (Arrays.asList("")); if (n == 1) return new ArrayList (Arrays.asList("1", "8", "0")); List list = helper(n - 2, max); List result = new ArrayList (); for (int i = 0; i < list.size(); i++) { String s = list.get(i); if (n != max) result.add("0" + s + "0"); result.add("1" + s + "1"); result.add("8" + s + "8"); result.add("6" + s + "9"); result.add("9" + s + "6"); } return result; } public int strobogrammaticInRange(String low, String high) { int count = 0; List result = new ArrayList (); for (int i = low.length(); i <= high.length(); i++) { result.addAll(helper(i, i)); } for (String str : result) { if (str.length() == low.length() && str.compareTo(low) < 0 || str.length() == high.length() && str.compareTo(high) > 0) continue; count++; } return count; } }
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摘要:题目链接这题和都可以做,一种思路就是从中间开始往两边延伸,每次有种可能性和,其中开头处不能是。可以加或者用优化。 247. Strobogrammatic Number II 题目链接:https://leetcode.com/problems... 这题recursion和iteration都可以做,一种思路就是从中间开始往两边延伸,每次c[i-k], c[i+k]有5种可能性: (...
Problem 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. Example: Input: n = 2Output...
Problem 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 represent...
Problem 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 o...
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