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.
- 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:
1 // 2WA, 1AC, fair enough~ 2 // #define MY_MAIN 3 // A line other than the last line might contain only one word. What should you do in this case? 4 // In this case, that line should be left-justified. 5 #include <cstdio> 6 #include <string> 7 #include <vector> 8 using namespace std; 9 10 class Solution { 11 public: 12 vector<string> fullJustify(vector<string> &words, int L) { 13 int ll, rr; 14 int i, j; 15 int wc = (int)words.size(); 16 int total_len; 17 int space_len1, space_len2; 18 int space_count1, space_count2; 19 int len; 20 21 result.clear(); 22 // 1WA here, special case is dealt with here. 23 if (wc == 1) { 24 s[0] = 0; 25 sprintf(s, "%s", words[0].c_str()); 26 len = (int)words[0].length(); 27 while (len < L) { 28 s[len++] = ' '; 29 } 30 s[len] = 0; 31 result.push_back(string(s)); 32 return result; 33 } 34 35 ll = 0; 36 while (ll < wc) { 37 rr = ll + 1; 38 total_len = (int)words[ll].length(); 39 while (rr < wc && total_len + 1 + (int)words[rr].length() <= L) { 40 total_len += ((int)words[rr].length() + 1); 41 ++rr; 42 } 43 44 if (rr < wc) { 45 // the total length exceeds L, more lines to be expected 46 // fully-justified 47 total_len -= (rr - ll - 1); 48 if (rr - ll > 1) { 49 space_count1 = (L - total_len) % (rr - ll - 1); 50 space_count2 = (rr - ll - 1) - space_count1; 51 space_len1 = (L - total_len) / (rr - ll - 1) + 1; 52 space_len2 = (L - total_len) / (rr - ll - 1); 53 s[0] = 0; 54 sprintf(s, "%s", words[ll].c_str()); 55 len = (int)words[ll].length(); 56 for (i = 0; i < space_count1; ++i) { 57 for (j = 0; j < space_len1; ++j) { 58 s[len++] = ' '; 59 } 60 s[len] = 0; 61 sprintf(s + len, "%s", words[ll + 1 + i].c_str()); 62 len += words[ll + 1 + i].length(); 63 } 64 for (i = 0; i < space_count2; ++i) { 65 for (j = 0; j < space_len2; ++j) { 66 s[len++] = ' '; 67 } 68 s[len] = 0; 69 sprintf(s + len, "%s", words[ll + 1 + space_count1 + i].c_str()); 70 len += words[ll + 1 + space_count1 + i].length(); 71 } 72 } else { 73 // only one word on this line, and not the last line 74 // left-justified 75 s[0] = 0; 76 sprintf(s, "%s", words[ll].c_str()); 77 for (i = (int)words[ll].length(); i < L; ++i) { 78 s[i] = ' '; 79 } 80 s[i] = 0; 81 } 82 } else { 83 // the last line is reached 84 // left-justified 85 s[0] = 0; 86 sprintf(s, "%s", words[ll].c_str()); 87 len = (int)words[ll].length(); 88 for (i = ll + 1; i < rr; ++i) { 89 s[len++] = ' '; 90 s[len] = 0; 91 sprintf(s + len, "%s", words[i].c_str()); 92 len += (int)words[i].length(); 93 } 94 // 1WA here, the last line must be filled with spaces. 95 while (len < L) { 96 s[len++] = ' '; 97 } 98 s[len] = 0; 99 } 100 result.push_back(string(s)); 101 ll = rr; 102 } 103 104 return result; 105 } 106 private: 107 char s[10000]; 108 vector<string> result; 109 }; 110 111 #ifdef MY_MAIN 112 int main() 113 { 114 int i; 115 int n; 116 vector<string> words; 117 int L; 118 char s[1000]; 119 Solution solution; 120 vector<string> result; 121 122 while (scanf("%d", &n) == 1 && n) { 123 scanf("%d", &L); 124 words.clear(); 125 for (i = 0; i < n; ++i) { 126 scanf("%s", s); 127 words.push_back(string(s)); 128 } 129 result = solution.fullJustify(words, L); 130 for (i = 0; i < (int)result.size(); ++i) { 131 printf("%d: %s ", i + 1, result[i].c_str()); 132 } 133 } 134 135 return 0; 136 } 137 #endif