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随便问一下我的实习经历, 讲了一分钟就直接coding了。 isAngram 两种方法及其变体。.鏈枃鍘熷垱鑷1point3acres璁哄潧
// Ransom Note
// magazine = "cdhjkscjgcdjkshvfhjgdu..." "themet"
// note = "Meet me somewhere" "meet" a-z A-Z 0-9

boolean canMakeRansomNote(String magazine, String note) {
//assume the input is not null
Map<Character, Frequence> map = new HashMap<>();
for(int i = 0; i < note.length(); i++){
char c = note.charAt(i);. from: 1point3acres.com/bbs
if(isValid(c)) //ignore any space or illegal chars{
int freq = map.containsKey(c) ? map.get(c)+1 : 1;. 1point 3acres 璁哄潧
map.put(c, freq);
}
}

for(int i = 0; i < magazine.length(); i++){
char c = magazine.charAt(i);. 1point 3acres 璁哄潧
if(isValid(c) && map.containsKey(c){ //care about the chars occured in notes.
int newFreq = map.get(c)-1; //found a char which is contained in notes
if(newFreq <= 0){
map.remove(c);
if(map.isEmpty())
return true;
}
else
map.put(c, newFreq);


}

return false;. 鐗涗汉浜戦泦,涓浜╀笁鍒嗗湴
}
}

public boolean isValid(char c){ //c is Uni-8, int = 65 ('A')
if(c >= 'a' && c <= 'z' || c >= 'A' && c <= 'Z' || c >= '0' && c <= '9')
return true;
return false;-google 1point3acres
}

// "abc", "cab" -> true "[a, b , c] [a, b, c]
// "abc", "def" -> false-google 1point3acres


void isAnagram(String string1, String string2) {
//first idea: sort them and then compare if it is equals.1point3acres缃
//assume input is not null
if(string1.length() != string2.length())
return false;

Map<Character, Integer> map = new HashMap<>();. 1point3acres.com/bbs
//iterate string1 and store it in map
for(int i = 0; i < string1.length(); i++){
char c = string1.charAt(i);
int freq = map.containsKey(c) ? map.get(c)+1 : 1;
map.put(c, freq);
}

//iterate the other string to check
for(int i = 0; i < string2.length(); i++){
char c = string2.charAt(i);
if(!map.containsKey(c)).鐣欏璁哄潧-涓浜-涓夊垎鍦
return false;. 鐗涗汉浜戦泦,涓浜╀笁鍒嗗湴
//decrease freq
int newFreq = map.get(c)-1;. 涓浜-涓夊垎-鍦帮紝鐙鍙戝竷
if(newFreq < 0)
return false;

map.put(c, newFreq);
}. From 1point 3acres bbs
. 1point3acres.com/bbs
return true;
}


. visit 1point3acres.com for more.

void isAnagram(String string1, String string2) {
//first idea: sort them and then compare if it is equals
//assume input is not null 鏉ユ簮涓浜.涓夊垎鍦拌鍧.
if(string1.length() != string2.length())
return false;

char[] S1 = string1.toCharArray();
char[] S2 = string2.toCharArray()

Arrays.sort(S1);.鏈枃鍘熷垱鑷1point3acres璁哄潧
Arrays.sort(s2);

return isEqual(S1, S2); // Arrays.toStint(s1).equals(Arrays.toString(S2));
}. From 1point 3acres bbs

public boolean isEqual(char[] A, char[] B){
//cornet case
if(A == null && B == null)
return true;
.1point3acres缃
//only one of them is null
if(A == null || B == null)
returnf false;

for(int i = 0; i < A.length; i++){. 1point3acres.com/bbs
if(A != B)
return false;
}. 1point3acres.com/bbs

return true;
}


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