LeetCode - 383. Ransom Note

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。