LeetCode:383. Ransom Note

eetCode:383. Ransom Note

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.

题目的意思就是给你两个字符串,判断第一个字符串ransom能不能由第二个字符串magazines里面的字符构成,第二个字符里的每个字符只能使用一次。(假设只包含小写字母)

  1. 那么首先想到的就是穷举法了,两层遍历即可,也就是下面的 canConstruct 方法了。

  2. 再思考一下,有个小技巧。这里只要判断两个字符串中每个字母出现的次数即可解决问题。对于某个字母来说,如果第ransom字符串中出现的次数比magazines中出现的次数要多,很显然就直接返回false了。

看一下 canConstructBetter 方法的实现:

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import java.util.List;
import java.util.ArrayList;
public class Solution {

public boolean canConstructBetter(String ransomNote, String magazine) {
if (ransomNote.equals(magazine) || "".equals(ransomNote)) {
return true;
}
if ("".equals(magazine) && !"".equals(ransomNote)) {
return false;
}

int[] ransomNoteCharCount = new int[26];
int[] magazineCharCount = new int[26];

for(int i = 0 ; i < ransomNote.length() ; i++) {
ransomNoteCharCount[ransomNote.charAt(i)-'a']++;
}

for(int i = 0 ; i < magazine.length() ; i++) {
magazineCharCount[magazine.charAt(i)-'a']++;
}

for(int i = 0 ; i < 26 ; i++) {
if(ransomNoteCharCount[i] > magazineCharCount[i] ) {
return false;
}
}
return true;
}

public boolean canConstruct(String ransomNote, String magazine) {
if (ransomNote.equals(magazine)) {
return true;
}
if ("".equals(ransomNote)) {
return true;
}
List<Character> ransomNodeChars = new ArrayList<>();
List<Character> magazineChars = new ArrayList<>();
for(int i = 0 ; i < ransomNote.length() ; i++) {
ransomNodeChars.add(ransomNote.charAt(i));
}
this.printList(ransomNodeChars);
for(int j = 0 ; j < magazine.length() ; j++) {
magazineChars.add(magazine.charAt(j));
}
boolean flag = false;
Character c = null;
for(int m = 0 ; m < ransomNote.length() ; m++) {
c = ransomNodeChars.get(m);
flag = false;
for(int n = 0 ; n < magazineChars.size() ; n++) {
if(magazineChars.get(n) == c) {
flag = true;
magazineChars.remove(n);
break;
}
}
if (flag == false) {
break;
}
}
return flag;
}
}

THE END

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