leetcode Weekly Contest 77 806. Number of Lines To Write String

806. Number of Lines To Write String

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·       User Accepted:1585

·       User Tried:1647

·       Total Accepted:1607

·       Total Submissions:2257

·       Difficulty:Easy

We are to write the letters of a given string S, fromleft to right into lines. Each line has maximum width 100 units, and if writinga letter would cause the width of the line to exceed 100 units, it is writtenon the next line. We are given an array widths, anarray where widths[0] is the width of 'a', widths[1] is the width of 'b', ...,and widths[25] is the width of 'z'.

Now answer two questions: how many lines have at leastone character from S, andwhat is the width used by the last such line? Return your answer as an integerlist of length 2.


Example :

Input:

widths =[10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10]

S = "abcdefghijklmnopqrstuvwxyz"

Output: [3, 60]

Explanation:

All letters have the same length of 10. To write all 26letters,

we need two full lines and one line with 60 units.

Example :

Input:

widths = [4,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10]

S = "bbbcccdddaaa"

Output: [2, 4]

Explanation:

All letters except 'a' have the same length of 10, and

"bbbcccdddaa" will cover 9 * 10 + 2 * 4 = 98 units.

For the last 'a', it is written on the second line because

there is only 2 units left in the first line.

So the answer is 2 lines, plus 4 units in the second line.

 

Note:

·       The length of S willbe in the range [1, 1000].

·       S willonly contain lowercase letters.

·       widths is anarray of length 26.

·       widths[i] willbe in the range of [2, 10].

这一题很简单,给你一个字符串,然后告诉你每一个字符有多长,告诉你每行最多多少个字符,问你一行一行排下来,一共需要多少行,以及最后一行有多少个字符

直接按照排列的规律来排列字符就能够得到最终的结果

class Solution {
public:
 vector<int> numberOfLines(vector<int>& widths, string S) {
        int theLength=0;
        int theLengthCount=1;
        int theLastLength=0;
        for(int i=0;i<(int)S.size();i++)
        {
        	if(theLength+widths[S[i]-'a']>100)
        	{
	        	theLastLength=theLength=widths[S[i]-'a'];
        		theLengthCount++;
	        }
	        else
	        {
        		theLength+=widths[S[i]-'a'];
        		theLastLength=theLength;
        	}
        }
        vector<int> result;
        result.push_back(theLengthCount);  result.push_back(theLastLength);
        return result;
}
};

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