[Leetcode] 68. Text Justification Problem Solving report

Source: Internet
Author: User

Title Link: https://leetcode.com/problems/text-justification/

Given an array of words and a length L, format the text such, all line have exactly L characters and is fully (left and right) justified.

You should pack your words in a greedy approach; That's, pack as many words as you can on each line. Pad extra spaces when necessary so, each line has ‘ ‘ exactly L characters.

Extra spaces between words should be distributed as evenly as possible. If the number of spaces on a line does not divide evenly between words, the empty slots on the left would be assigned more SP Aces than the slots on the right.

For the last line of text, it should being justified and no extra space is inserted between words.

For example,
words:["This", "is", "an", "example", "of", "text", "justification."]
L: 16 .

Return the formatted lines as:

[   "This    was an",   "example  of text", "   justification.  "]

Note: Each word was guaranteed not to exceed L in length.

Idea: is a string processing of the topic, not difficult, but it is difficult to write more elegant.

Two code as follows, the first one is my own, the code is longer, 0ms. The second one is the reference of others, 4ms, but more beautiful.

The second method is very clever to find the unbalanced space, is to use the position and the remainder of the relationship to the left to allocate extra space.

The code is as follows:

Class Solution {public:vector<string> Fulljustify (vector<string>& words, int maxWidth) {int cn        t = 0, left = 0;        vector<string> result;            for (int i =0; i< words.size (); i++) {cnt + = words[i].size (); if (Cnt+i-left > MaxWidth | | i+1==words.size ()) {if (Cnt+i-left > MaxWidth) cnt-= words[i-                -].size ();                String str = Words[left];                    for (int j = left+1; j<= i; j + +) {int m = maxwidth-cnt, n = i-left;                    if (i+1==words.size ()) str + = "";                    else Str.append (m/n + (j-left-1<m% n), ");                str + = Words[j];                } str.append (Maxwidth-str.size (), ");                Result.push_back (str);            left = i+1, cnt = 0;    }} return result; }};

Class Solution {public:    vector<string> fulljustify (vector<string>& words, int maxWidth) {        vector<string> result;        for (int i=0, m, N; i< words.size (); i+=m)        {for            (M=0,n=0;i+m<words.size () &&m+n+words[i+m].size ( ) <=maxWidth; m++)                 n+= words[i+m].size ();            String str = words[i];            for (int j = 0; J < M-1; J + +)            {                if (i+m >= words.size ()) str+= "";                else                    Str.append ((maxwidth-n)/(m-1) + (j< (maxwidth-n)% (m-1)), ');                str + = words[i+j+1];            }            Str.append (Maxwidth-str.size (), ');            Result.push_back (str);        }        return result;}    ;
Second type of reference: Https://leetcode.com/discuss/13610/share-my-concise-c-solution-less-than-20-lines

[Leetcode] 68. Text Justification Problem Solving report

Related Keywords:

Contact Us

The content source of this page is from Internet, which doesn't represent Alibaba Cloud's opinion; products and services mentioned on that page don't have any relationship with Alibaba Cloud. If the content of the page makes you feel confusing, please write us an email, we will handle the problem within 5 days after receiving your email.

If you find any instances of plagiarism from the community, please send an email to: info-contact@alibabacloud.com and provide relevant evidence. A staff member will contact you within 5 working days.

A Free Trial That Lets You Build Big!

Start building with 50+ products and up to 12 months usage for Elastic Compute Service

• Sales Support

1 on 1 presale consultation

• After-Sales Support

24/7 Technical Support 6 Free Tickets per Quarter Faster Response

• Alibaba Cloud offers highly flexible support services tailored to meet your exact needs.