05 August 2008
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.
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canConstruct("a", "b") -> false
canConstruct("aa", "ab") -> false
canConstruct("aa", "aab") -> true
赎金条，看一个magazines来的字符串是否可以组成赎金的note的字符串，每个字符只能用一次。用HashMap来统计次数，然后把赎金条的信息遍历看是否够用即可。
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class Solution {
public boolean canConstruct(String ransomNote, String magazine) {
Map<Character, Integer> map = new HashMap<>();
for (char ch : magazine.toCharArray()) {
int newCount = map.getOrDefault(ch, 0) + 1;
map.put(ch, newCount);
}
for (char ch : ransomNote.toCharArray()) {
if (!map.containsKey(ch)) {//不包含的字符直接为false
return false;
}
int newCount = map.get(ch) - 1;
if (newCount < 0) {//ransomNote字符过多返回false
return false;
}
map.put(ch, newCount);//map.get(ch) = newCount;
}
return true;
}
}
用array，类似Valid Anagram的解法，用一个数组来统计字符出现的次数。
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class Solution {
public boolean canConstruct(String ransomNote, String magazine) {
int[] chs = new int[26];
for (int i = 0; i < ransomNote.length(); i++) {
chs[ransomNote.charAt(i) - 'a']--;
}
for (int i = 0; i < magazine.length(); i++) {
chs[magazine.charAt(i) - 'a']++;
}
for (int ch : chs) {
if (ch < 0) {
return false;
}
}
return true;
}
}