LeetCode

Text Justification

2014.2.10 02:42

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

You should pack your words in a greedy approach; that is, pack as many words as you can in each line. Pad extra spaces ' ' when necessary so that each line has exactly Lcharacters.

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

For the last line of text, it should be left 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    is    an",
   "example  of text",
   "justification.  "
]

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

click to show corner cases.

Corner Cases:

• A line other than the last line might contain only one word. What should you do in this case?
  In this case, that line should be left-justified.

Solution:

　　The problem is not so difficult, but you have to carefully tackle several corner cases. Refer to the hint and think for yourself.

　　Hope you can solve it with one submission only. Pity I was too careless then.

　　Total time complexity is O(n), where n is the total number of words. Space complexity is O(total length of all words).

Accepted code:

// 2WA, 1AC, fair enough~
// #define MY_MAIN
// A line other than the last line might contain only one word. What should you do in this case?
// In this case, that line should be left-justified.
#include <cstdio>
#include <string>
#include <vector>
using namespace std;

class Solution {
public:
    vector<string> fullJustify(vector<string> &words, int L) {
        int ll, rr;
        int i, j;
        int wc = (int)words.size();
        int total_len;
        int space_len1, space_len2;
        int space_count1, space_count2;
        int len;
        
        result.clear();
        // 1WA here, special case is dealt with here.
        if (wc == 1) {
            s[0] = 0;
            sprintf(s, "%s", words[0].c_str());
            len = (int)words[0].length();
            while (len < L) {
                s[len++] = ' ';
            }
            s[len] = 0;
            result.push_back(string(s));
            return result;
        }
        
        ll = 0;
        while (ll < wc) {
            rr = ll + 1;
            total_len = (int)words[ll].length();
            while (rr < wc && total_len + 1 + (int)words[rr].length() <= L) {
                total_len += ((int)words[rr].length() + 1);
                ++rr;
            }
            
            if (rr < wc) {
                // the total length exceeds L, more lines to be expected
                // fully-justified
                total_len -= (rr - ll - 1);
                if (rr - ll > 1) {
                    space_count1 = (L - total_len) % (rr - ll - 1);
                    space_count2 = (rr - ll - 1) - space_count1;
                    space_len1 = (L - total_len) / (rr - ll - 1) + 1;
                    space_len2 = (L - total_len) / (rr - ll - 1);
                    s[0] = 0;
                    sprintf(s, "%s", words[ll].c_str());
                    len = (int)words[ll].length();
                    for (i = 0; i < space_count1; ++i) {
                        for (j = 0; j < space_len1; ++j) {
                            s[len++] = ' ';
                        }
                        s[len] = 0;
                        sprintf(s + len, "%s", words[ll + 1 + i].c_str());
                        len += words[ll + 1 + i].length();
                    }
                    for (i = 0; i < space_count2; ++i) {
                        for (j = 0; j < space_len2; ++j) {
                            s[len++] = ' ';
                        }
                        s[len] = 0;
                        sprintf(s + len, "%s", words[ll + 1 + space_count1 + i].c_str());
                        len += words[ll + 1 + space_count1 + i].length();
                    }
                } else {
                    // only one word on this line, and not the last line
                    // left-justified
                    s[0] = 0;
                    sprintf(s, "%s", words[ll].c_str());
                    for (i = (int)words[ll].length(); i < L; ++i) {
                        s[i] = ' ';
                    }
                    s[i] = 0;
                }
            } else {
                // the last line is reached
                // left-justified
                s[0] = 0;
                sprintf(s, "%s", words[ll].c_str());
                len = (int)words[ll].length();
                for (i = ll + 1; i < rr; ++i) {
                    s[len++] = ' ';
                    s[len] = 0;
                    sprintf(s + len, "%s", words[i].c_str());
                    len += (int)words[i].length();
                }
                // 1WA here, the last line must be filled with spaces.
                while (len < L) {
                    s[len++] = ' ';
                }
                s[len] = 0;
            }
            result.push_back(string(s));
            ll = rr;
        }
        
        return result;
    }
private:
    char s[10000];
    vector<string> result;
};

#ifdef MY_MAIN
int main()
{
    int i;
    int n;
    vector<string> words;
    int L;
    char s[1000];
    Solution solution;
    vector<string> result;
    
    while (scanf("%d", &n) == 1 && n) {
        scanf("%d", &L);
        words.clear();
        for (i = 0; i < n; ++i) {
            scanf("%s", s);
            words.push_back(string(s));
        }
        result = solution.fullJustify(words, L);
        for (i = 0; i < (int)result.size(); ++i) {
            printf("%d: %s
", i + 1, result[i].c_str());
        }
    }
    
    return 0;
}
#endif