赎金信(Ransom Not)

给定一个赎金信 (ransom) 字符串和一个杂志(magazine)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。如果可以构成，返回 true ；否则返回 false。

(题目说明：为了不暴露赎金信字迹，要从杂志上搜索各个需要的字母，组成单词来表达意思。杂志字符串中的每个字符只能在赎金信字符串中使用一次。)

示例 1：

输入：ransomNote = "a", magazine = "b"
输出：false

示例 2：

输入：ransomNote = "aa", magazine = "ab"
输出：false

示例 3：

输入：ransomNote = "aa", magazine = "aab"
输出：true

提示：

  • 你可以假设两个字符串均只含有小写字母。
bool canConstruct(char * ransomNote, char * magazine){
    int ransomNoteSize = strlen(ransomNote);
    int magazineSize = strlen(magazine);

    for(int i =0;i< ransomNoteSize; i++){
        bool isFound = false;
        for(int j = 0; j < magazineSize; j++) {
            if(magazine[j] == ransomNote[i]) {
                isFound = true;
                magazine[j] = '-';
                break;
            }
        }
        if (!isFound){
            return false;
        }
    }
    return true;
}

找到一个把报纸上的nage 字母挖掉

另一种思路，看看报纸上的内容可不可以组成想要的内容

bool canConstruct(char * ransomNote, char * magazine){
    int ransomNoteSize = strlen(ransomNote);
    int magazineSize = strlen(magazine);

    for(int i =0;i< ransomNoteSize; i++){
        bool isFound = false;
        for(int j = i; j < magazineSize; j++) {
            if(magazine[j] == ransomNote[i]) {
                isFound = true;
                char temp = magazine[j];
                magazine[j] = magazine[i];
                magazine[i] =temp;
                break;
            }
        }
        if (!isFound){
            return false;
        }
    }
    return true;
}

bool canConstruct(char * ransomNote, char * magazine){
    int ransomNoteSize = strlen(ransomNote);
    int magazineSize = strlen(magazine);
    // 这个字符在左边出现的次数小于在右边出现的次数

    for(int i =0;i< ransomNoteSize; i++){

        int countRansomNote = 0;
        for (int j = 0; j< ransomNoteSize; j++){
            if(magazine[j] == ransomNote[i]){
                countRansomNote++;
            }
        }

        int countMagazine = 0;
        for (int j =0; j< magazineSize; j++){
            if(magazine[j] == ransomNote[i]){
                countMagazine++;
            }
        }
        if (countRansomNote > countMagazine){
            return false;
        }
    }
    return true;
}

但是这时候还是有O(n^2) 的时间复杂度，这时候大家可能已经可以想到 O(n) 的解法了

bool canConstruct(char * ransomNote, char * magazine){
    int ransomNoteSize = strlen(ransomNote);
    int magazineSize = strlen(magazine);
    // 这个字符在左边出现的次数小于在右边出现的次数

    int countRansomNote['z'-'a'+1] = {0};
    for (int i =0; i < ransomNoteSize; i++){
        countRansomNote[ransomNote[i]-'a']++;
    }

    int countMagazine['z'-'a'+1] = {0};
    for (int i =0; i < magazineSize; i++){
        countMagazine[magazine[i]-'a']++;
    }

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

甚至我们只要一个数组就可以了

bool canConstruct(char * ransomNote, char * magazine){
    int ransomNoteSize = strlen(ransomNote);
    int magazineSize = strlen(magazine);
    // 这个字符在左边出现的次数小于在右边出现的次数

    int countDiff['z'-'a'+1] = {0};
    for (int i =0; i < ransomNoteSize; i++){
        countDiff[ransomNote[i]-'a']++;
    }
    for (int i =0; i < magazineSize; i++){
        countDiff[magazine[i]-'a']--;
    }

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

也可以排完序后来比比看