非常好的一个思想就是,用int[26]来存储字典里26个英文字母的出现次数map[c - ‘a’]++;然后逐次比较两个map,一个单词循环26次计较(subMap[j] == 0就不用比较了)。
public int countCharacters1(String[] word,String chars){
int[] map = new int[26];
for (char c : chars.toCharArray()){
map[c - 'a']++;
}
int res = 0;
for (int i = 0; i < word.length; i++) {
int[] subMap = new int[26];
for (char c : word[i].toCharArray()) {
subMap[c - 'a']++;
}
for (int j = 0; j < 26; j++) {
if (map[j] < subMap[j]){
break;
}
if (j == 25){
res += word[i].length();
}
}
}
return res;
}
自己写的费时费力,用的list存,每次拼写结束以后,恢复当前的list的初始的list。每次拼写需要判断存不存在,不存在直接break; 字母存在的话,就继续循环比较。
public int countCharacters(String[] words, String chars) {
if (chars == null || chars.length() == 0){
return 0;
}
char[] array = chars.toCharArray();
ArrayList<Character> list = new ArrayList<>();
for (int i = 0; i < array.length; i++) {
list.add(array[i]);
}
ArrayList<Character> subList = new ArrayList<>();
int count = 0;
for (int i = 0; i < words.length; i++) {
subList.clear();
subList.addAll(list);
for (int j = 0; j < words[i].length(); j++) {
if (subList.contains(words[i].charAt(j))){
subList.remove((Character) words[i].charAt(j));
} else {
break;
}
if (j == words[i].length() - 1){
count += words[i].length();
}
}
}
return count;
}