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 }