給你兩個字符串：ransomNote?和?magazine?，判斷?ransomNote?能不能由?magazine?里面的字符構成。

如果可以，返回?true?；否則返回?false?。

magazine?中的每個字符只能在?ransomNote?中使用一次。

示例 1：

輸入：ransomNote = "a", magazine = "b"
輸出：false

示例 2：

輸入：ransomNote = "aa", magazine = "ab"
輸出：false

示例 3：

輸入：ransomNote = "aa", magazine = "aab"
輸出：true

提示：

• 1 <= ransomNote.length, magazine.length <= 105
• ransomNote?和?magazine?由小寫英文字母組成

?代碼：

bool canConstruct(char* ransomNote, char* magazine) { // leeCode 383.贖金信int len1 = strlen(ransomNote);int len2 = strlen(magazine);if (len1 > len2)return false;int ransomNoteNumArr[26]; // 統計字符串ransomNote中每個字母出現次數。如ransomNoteNumArr[0] 為a出現的次數memset(ransomNoteNumArr, 0, sizeof(int) * 26);int magazineNumArr[26]; // 統計字符串magazine中每個字母出現次數。如magazineNumArr[0] 為a出現的次數memset(magazineNumArr, 0, sizeof(int) * 26);for (int i = 0; i < len1; i++) {char c = ransomNote[i];int idx = c - 'a';ransomNoteNumArr[idx]++;}for (int i = 0; i < len2; i++) {char c = magazine[i];int idx = c - 'a';magazineNumArr[idx]++;}for (int i = 0; i < 26; i++) {if (ransomNoteNumArr[i] > magazineNumArr[i]) {return false;}}return true;
}

測試代碼：

void testLeeCode383() {const char* s1 = "aa";const char* s2 = "aab";char ransomNote[3];char magazine[4];memcpy(ransomNote, s1, sizeof(char) * 3);memcpy(magazine, s2, sizeof(char) * 4);printf("ransomNote: %s, magazine: %s\n", ransomNote, magazine);if (canConstruct(ransomNote, magazine)) {printf("true\n");}else {printf("false\n");}
}

打印結果：

測試ok。題外話，為啥這題標題叫贖金信？