给定一个单词数组和一个长度 maxWidth，重新排版单词，使其成为每行恰好有 maxWidth 个字符，且左右两端对齐的文本。
你应该使用“贪心算法”来放置给定的单词；也就是说，尽可能多地往每行中放置单词。必要时可用空格 ' ' 填充，使得每行恰好有 maxWidth 个字符。
要求尽可能均匀分配单词间的空格数量。如果某一行单词间的空格不能均匀分配，则左侧放置的空格数要多于右侧的空格数。
文本的最后一行应为左对齐，且单词之间不插入额外的空格。
说明:
- 单词是指由非空格字符组成的字符序列。
- 每个单词的长度大于 0，小于等于 maxWidth。
- 输入单词数组
words至少包含一个单词。
示例:
输入: words = ["This", "is", "an", "example", "of", "text", "justification."] maxWidth = 16 输出: [ "This is an", "example of text", "justification. " ]
示例 2:
输入:
words = ["What","must","be","acknowledgment","shall","be"]
maxWidth = 16
输出:
[
"What must be",
"acknowledgment ",
"shall be "
]
解释: 注意最后一行的格式应为 "shall be " 而不是 "shall be",
因为最后一行应为左对齐，而不是左右两端对齐。
第二行同样为左对齐，这是因为这行只包含一个单词。
示例 3:
输入: words = ["Science","is","what","we","understand","well","enough","to","explain", "to","a","computer.","Art","is","everything","else","we","do"] maxWidth = 20 输出: [ "Science is what we", "understand well", "enough to explain to", "a computer. Art is", "everything else we", "do " ]
1 #include "_000库函数.h"
2
3 class Solution {
4 public:
5 vector<string> fullJustify(vector<string>& words, int maxWidth) {
6 vector<string>res;
7 vector<string>lin;//一行单词的存放量
8 int n = 0;//一行字符的长度
9 for (int i = 0; i < words.size(); ++i) {
10 n += words[i].size();//单词长度
11 lin.push_back(words[i]);
12 if (n + lin.size() - 1 > maxWidth) {//长度超过了lin.size() - 1为需要加空格的长度
13 n -= words[i].size();
14 lin.pop_back();//最后一个单词放不进去
15 string str = "";
16 int num = (lin.size() - 1) > 0 ? (lin.size() - 1) : (maxWidth - n + 1);//用来防止只有一个单词的情况
17 int k = (maxWidth - n) % num;//如果无法均分，则左边的空格要更多
18 for (int j = 0; j < lin.size(); ++j) {
19 str += lin[j];
20 if (lin.size() == 1) {//只有一个单词
21 str.insert(str.end(), k, ' ');
22 break;
23 }
24 if (k - j > 0)str += " ";//左边的多加空格
25 if (j < lin.size() - 1)
26 str.insert(str.end(), (maxWidth - n) / num, ' ');
27 }
28 res.push_back(str);
29 --i;
30 n = 0;
31 lin.clear();
32 }
33 }
34 if (!lin.empty()) {//最后一点单词了
35 string str = "";
36 for (int j = 0; j < lin.size(); ++j) {
37 str += lin[j];
38 if (j < lin.size() - 1)
39 str += " ";//最后一行为左对齐，单词间只需要添加一个空格就行
40 }
41 str.insert(str.end(), (maxWidth - n - lin.size() + 1), ' ');//最后拿剩余的空格顶替
42 res.push_back(str);
43 }
44 return res;
45 }
46 };
47
48
49 //方法二，更简洁，但跟方法一的思想一样
50 class Solution {
51 public:
52 vector<string> fullJustify(vector<string> &words, int L) {
53 vector<string> res;
54 int i = 0;
55 while (i < words.size()) {
56 int j = i, len = 0;
57 while (j < words.size() && len + words[j].size() + j - i <= L) {
58 len += words[j++].size();
59 }
60 string out;
61 int space = L - len;
62 for (int k = i; k < j; ++k) {
63 out += words[k];
64 if (space > 0) {
65 int tmp;
66 if (j == words.size()) {
67 if (j - k == 1) tmp = space;
68 else tmp = 1;
69 }
70 else {
71 if (j - k - 1 > 0) {
72 if (space % (j - k - 1) == 0) tmp = space / (j - k - 1);
73 else tmp = space / (j - k - 1) + 1;
74 }
75 else tmp = space;
76 }
77 out.append(tmp, ' ');
78 space -= tmp;
79 }
80 }
81 res.push_back(out);
82 i = j;
83 }
84 return res;
85 }
86 };
87
88 void T068() {
89 Solution s;
90 vector<string>v;
91 v = { "aaaaaa","bbbbbb","This", "is", "an", "example", "of", "text", "justification." };
92 v = s.fullJustify(v, 16);
93 for (auto a : v)
94 cout << a << endl;
95 cout << endl;
96 v = { "What","must","be","acknowledgment","shall","be"};
97 v = s.fullJustify(v, 16);
98 for (auto a : v)
99 cout << a << endl;
100 cout << endl;
101 v = { "Science","is","what","we","understand","well","enough","to","explain",
102 "to","a","computer.","Art","is","everything","else","we","do" };
103 v = s.fullJustify(v, 20);
104 for (auto a : v)
105 cout << a << endl;
106 cout << endl;
107
108 }
来源：https://www.cnblogs.com/zzw1024/p/10696199.html