Leetcode刷题笔记6:ZigZag Conversion(C语言)

char* convert(char* s, int numRows) {
    char str[numRows+1][1000];
    int i=0,j=0,n,p=0,flag=0;//p记录现在填充到s的第几个字符了
    n=strlen(s);
    char *str2;
    str2=(char *)malloc((n+1)*sizeof(char));
    while(p<n)
    {
        if(numRows>1)flag=flag%(numRows-1);
        else flag=0;
        j++;//一列一列填入数字
       if(flag==0)//“之”的首末要全部填上
       {
           if(n-p>=numRows)//剩余的字符可以填满
           {
               for(i=1;i<=numRows;i++)
                   str[i][j]=s[p++];
           }
           else
           {
               int q=n-p+1;
               for(i=1;i<=q;i++)
                   str[i][j]=s[p++];
               for(i=q;i<=numRows;i++)
               {
                   str[i][j]=' ';
               }
           }
       }
        else//每一列只需要填一个字符
        {
            for(i=1;i<=numRows;i++)
            {
                if(i==(numRows-flag))
                {
                    str[i][j]=s[p++];
                }
                else
                {
                	str[i][j]=' ';
                } 
            }
        }
        flag++;
    }
    n=j;//共j列
    p=0;
    for(i=1;i<=numRows;i++)
    {
        for(j=1;j<=n;j++)
        {
             if(str[i][j]!=' ')
            {
                str2[p]=str[i][j];p++;
            }
        }
    }
    str2[p]='\0';
    return str2;
}

更优的sample:

char* convert(char* s, int numRows) {
	if (numRows == 1) return s;
	else if (numRows == 0) return NULL;

    int len = strlen(s);
    int i;
    int ind[10000];
    char ret[10000] = {0};
    //int *ind = (int*)malloc(sizeof(int) * len);
    //char *ret = (char*)malloc(sizeof(char) * (len+1));
    int mod = numRows*2-2;

    int index = 0;
    for (i = 0;i < numRows;i++){
    	int count = i;
    	if (i != 0 && i != numRows-1){
    		while(count<len){
    			ind[count] = index++;
    			count = count + mod - i - i;
    			if (count >= len) break;
    			ind[count] = index++;
    			count = count + i + i;    			
    		}
    	}
    	else {
    		while(count<len){
    			ind[count] = index++;
    			count += mod;
    		}
    	}
    }

    for (i = 0;i<len;i++){
    	ret[ind[i]] = s[i];
    }
    strcpy(s, ret);
    //free(ret);
    //free(ind);
    return s;
}

 

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转载自blog.csdn.net/weixin_41036461/article/details/89248122
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