2019-04-21

狗哥の情书

//这个题首先把源字符串做分割，把列表中的每个词也做分割，判断源字符串的分割能否被列表中单词的分割一一对应上。其实重点就是如何按照重复情况进行字符串分割。
int expressiveWords(char* S, char** words, int wordsSize) {
    int result = 0;
    for(int i = 0; i < wordsSize; i++)//比较words[i] 和 s
    {
        int a = 0, b = 0, yes = 1;
        while(words[i][a] != '\0' && S[b]!= '\0')
        {
            if(words[i][a] != S[b]) {yes = 0; break;}
            else 
            {
                int count_a = 0, count_b = 0;
                char ca = words[i][a], cb = S[b];
                while(words[i][a] == ca) {count_a++; a++;}
                while(S[b] == cb) {count_b++; b++;}
                if(count_a == count_b) continue;
                else if(count_b < count_a || count_b < 3) {yes = 0; break;}
            }
        }
        if (words[i][a] != '\0' ||  S[b]!= '\0') yes = 0;
        if(yes) result++;
    }
    return result;
}

提莫进阶攻略

//题目给了我们一个timeSeries， 和一个 duration，让我们找出中毒的total 时间。流程大概就是每次检查相邻的两个段是否有重合，如果有，则加上不重合的部分；否则就加上整个duration。
int findPoisonedDuration(int* timeSeries, int timeSeriesSize, int duration) {
    int total = 0;
    int i;
    
    if(timeSeriesSize>0){
        total = duration;
    }
    for(i=1; i<timeSeriesSize; i++){
            if(timeSeries[i] - timeSeries[i-1]>=duration){
                total += duration;
            }else{
                total += timeSeries[i] - timeSeries[i-1];
            }
    }

    return total;
}