383. Ransom Note

####QUESTION: 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.

Note: You may assume that both strings contain only lowercase letters.

canConstruct(“a”, “b”) -> false canConstruct(“aa”, “ab”) -> false canConstruct(“aa”, “aab”) -> true

####EXPLANATION: 首先将ransom的字符串分解成单个的，然后一个一个的去进行匹配，但是需要注意重复的需要从前一个匹配的结果处进行。
于是就想到了用map去记录上一个字符的位置，具体的看代码吧。

####SOLUTION:

public boolean canConstruct(String ransomNote, String magazine) {
    char[] chars = ransomNote.toCharArray();
    Map<Character,Integer> mapping = new HashMap<>();
    for (char alpha:chars
            ){
        if(mapping.containsKey(alpha)){
            Integer index = mapping.get(alpha);
            index = magazine.indexOf(alpha,index+1);
            if(index != -1){
                mapping.put(alpha,index);
            }else{
                return false;
            }
        }else{
            mapping.put(alpha,-1);
            int index = magazine.indexOf(alpha, 0);
            if(index!=-1){
                mapping.put(alpha,index);
            }else{
                return false;
            }
        }
    }
    return true;
}
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--> return <--