LeetCode - 383. Ransom Note
2017-02-16 22:24
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题目:
Given an arbitrary ransom note string and another string containing letters from all the magazines, write a function that will return true if the ransom note can be constructed from the magazines ; otherwise, it will return false.Each letter in the magazine string can only be used once in your ransom note.
题目解析:
即判断 ransomNote 是否可以通过magazine中的字母组成,其中 magazine 中的字母不考虑顺序,即只要有这个字母就行。注意的是 magazine 中每个字母只能用一次。思路与步骤:
两种思路
1. Comparing the character in ransomNote with the magazine directly. Obviously, nest loop has poor complexity. So I propose an easy-understand solution with Listwhich is faster than Map in sulution 2.2. Using a certain structure to store the count of every character. For example, usingarrayand Map. This solution is so interesting and clever. And
I also learn some basis of Java.
编程基础:
1. 循环的简介写法,不需要数组下标for (char c : str.toCharArray())
2. 数组下标
arr[c-'a']++;
3. Map 的学习
Java程序实现:
方法1:直接比较 —— List
public class Solution { public boolean canConstruct(String ransomNote, String magazine) { char[] rArray = ransomNote.toCharArray(); char[] mArray = magazine.toCharArray(); List<Character> rlist = new ArrayList<Character>(); for(int i = 0; i<magazine.length(); i++) rlist.add(mArray[i]); for(int i = 0; i<ransomNote.length(); i++){ if(rlist.contains(rArray[i])) rlist.remove(rlist.indexOf(rArray[i])); else return false; } return true; } }
方法2:
—— Array
public class Solution { public boolean canConstruct(String ransomNote, String magazine) { int[] arr = new int[26]; // method-1 for(char c: magazine.toCharArray()) arr[c-'a']++; for(char c: ransomNote.toCharArray()) if(--arr[c-'a'] < 0) return false; /* // method-2 for(int i = 0; i<magazine.length(); i++) arr[magazine.charAt(i)-'a']++; for(int i = 0; i<ransomNote.length(); i++) if(--arr[ransomNote.charAt(i)-'a'] < 0) return false; */ return true; } }
—— Map
public class Solution { public boolean canConstruct(String ransomNote, String magazine) { // method-3 Map<Character, Integer> map = new HashMap<>(); for(char c: magazine.toCharArray()){ int count = map.containsKey(c) ? map.get(c)+1 : 1; map.put(c, count); } for(char c: ransomNote.toCharArray()){ int newCount = map.containsKey(c) ? map.get(c)-1 : -1; if(newCount == -1) return false; map.put(c, newCount); } return true; } }
方法2 中method-1 最快,其次method-2,用Map 比List 还慢。
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