Given an array of words and a width maxWidth, format the text such that each line has exactly maxWidth 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 maxWidth

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.

Note:

• A word is defined as a character sequence consisting of non-space characters only.
• Each word‘s length is guaranteed to be greater than 0 and not exceed maxWidth.
• The input array words contains at least one word.

Example 1:

Input:
words = ["This", "is", "an", "example", "of", "text", "justification."]
maxWidth = 16
Output:
[
   "This    is    an",
   "example  of text",
   "justification.  "
]
 

Example 2:

Input:
words = ["What","must","be","acknowledgment","shall","be"]
maxWidth = 16
Output:
[
  "What   must   be",
  "acknowledgment  ",
  "shall be        "
]
Explanation: Note that the last line is "shall be    " instead of "shall     be",
             because the last line must be left-justified instead of fully-justified.
             Note that the second line is also left-justified becase it contains only one word.

Example 3:

Input:
words = ["Science","is","what","we","understand","well","enough","to","explain",
         "to","a","computer.","Art","is","everything","else","we","do"]
maxWidth = 20
Output:
[
  "Science  is  what we",
  "understand      well",
  "enough to explain to",
  "a  computer.  Art is",
  "everything  else  we",
  "do                  "
]

題意

確定最大行字符數為max，將一段文本以每行max的標準左右對齊

對於每行來說，單詞之間可以用盡量均衡的空格隔開，如果空格實在均衡不了的話就左邊的空格數比右邊的大

如果一行只有一個單詞，左對齊，右邊都用空格補齊

最後一行單詞之間的空格數只能是1

題解

一開始挺慢的

 1 class Solution {
 2 public:
 3     vector<string> fullJustify(vector<string> &words, int L) {
 4         vector<string> res;
 5         int i = 0;
 6         while (i < words.size()) {
 7             int j = i, len = 0;
 8             while (j < words.size() && len + words[j].size() + j - i <= L) {
 9                 len += words[j++].size();
10             }
11             string out;
12             int space = L - len;
13             for (int k = i; k < j; ++k) {
14                 out += words[k];
15                 if (space > 0) {
16                     int tmp;
17                     if (j == words.size()) { 
18                         if (j - k == 1) tmp = space;
19                         else tmp = 1;
20                     } else {
21                         if (j - k - 1 > 0) {
22                             if (space % (j - k - 1) == 0) tmp = space / (j - k - 1);
23                             else tmp = space / (j - k - 1) + 1;
24                         } else tmp = space;
25                     }
26                     out.append(tmp, ‘ ‘);
27                     space -= tmp;
28                 }
29             }
30             res.push_back(out);
31             i = j;
32         }
33         return res;
34     }
35 };

後來循環加空格的操作用append函數代替後就快了一半

 1 class Solution {
 2 public:
 3     vector<string> fullJustify(vector<string>& words, int maxWidth) {
 4         int n = words.size(), i = 0, sum = -1, last = 0;
 5         vector<string>res;
 6         while (i < n) {
 7             int newsum = 1 + words[i].length() + sum;
 8             if (newsum > maxWidth && i-last>1) {
 9                 int lack = maxWidth - sum;
10                 int every = lack / (i - last - 1), extra = lack % (i - last - 1), j;
11                 string ans = "";
12                 for (j = 0; j < extra; j++) {
13                     ans += words[last + j];
14                     if (j != i - last - 1)
15                         ans.append(every + 2, ‘ ‘);
16                 }
17                 for (; j < (i-last); j++) {
18                     ans += words[last + j];
19                     if (j != i - last - 1) 
20                         ans.append(every + 1, ‘ ‘);
21                 }
22                 sum = -1; last = i;
23                 res.push_back(ans);
24             }
25             else if (newsum > maxWidth) {
26                 string ans = "";
27                 ans += words[last];
28                 int lack = maxWidth - ans.length();
29                 ans.append(lack, ‘ ‘);
30                 sum = -1; last = i;
31                 res.push_back(ans);
32             }
33             else if (i == n - 1) {
34                 string ans = "";
35                 ans += words[last];
36                 for (int j = last + 1; j <= i; j++)
37                     ans += " " + words[j];
38                 int lack = maxWidth - ans.length();
39                 ans.append(lack, ‘ ‘);
40                 res.push_back(ans);
41                 i++;
42             }
43             else {
44                 sum = newsum;
45                 i++;
46             }
47         }
48         return res;
49     }
50 };

19.2.12 [LeetCode 68] Text Justification