标签：style blog http io ar color os sp java
Text Justification
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 L characters.
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.
SOLUTION 1:
递归解答：
每一次只处理一行，其它行交给下一级递归来处理。
思路是：
1. 先贪心法取得我们这一行可以放多少单词。
2. 在每个单词后面计算空格1个，但最后一个的要加回来。因为最后一个单词不需要加空格在右边。
3. 余下的空格数，平均分配给所有的interval。
4. 如果不能取整，多出的从左到右分配给那些interval。
5. 如果是1个单词，或是最后一行，在最后补空格。
其实整个题目不算难。但很繁杂，起码提交了8，9次才算过。
1 public class Solution { 2 public List<String> fullJustify(String[] words, int L) { 3 List<String> ret = new ArrayList<String>(); 4 5 if (words == null) { 6 return ret; 7 } 8 9 rec(words, L, 0, ret); 10 return ret; 11 } 12 13 public static void rec(String[] words, int L, int index, List<String> list) { 14 int len = words.length; 15 if (index >= len) { 16 return; 17 } 18 19 int LenTmp = L; 20 21 int end = index; 22 for (int i = index; i < len && words[i].length() <= L; i++) { 23 L -= words[i].length(); 24 25 // the space follow the word. 26 L--; 27 end = i; 28 } 29 30 // 最后一个空格收回 31 L++; 32 33 // Count how many words do we have. 34 int num = end - index + 1; 35 36 int extraSpace = 0; 37 int firstExtra = 0; 38 39 // 单词数大于1，才需要分配，否则所有的空格加到最后即可 40 if (num > 1) { 41 extraSpace = L / (num - 1); 42 // 首单词后多跟的空格 43 firstExtra = L % (num - 1); 44 } 45 46 StringBuilder sb = new StringBuilder(); 47 for (int i = index; i <= end; i++) { 48 sb.append(words[i]); 49 50 // The space following every word. 51 if (i != end) { 52 sb.append(‘ ‘); 53 } 54 55 // 不是最后一行 56 if (end != len - 1) { 57 // The first words. 58 if (firstExtra > 0) { 59 sb.append(‘ ‘); 60 firstExtra--; 61 } 62 63 // 最后一个单词后面不需要再加空格 64 if (i == end) { 65 break; 66 } 67 68 // 每个单词后的额外空格 69 int cnt = extraSpace; 70 while (cnt > 0) { 71 sb.append(‘ ‘); 72 cnt--; 73 } 74 } 75 } 76 77 // 最后一行的尾部的空格 78 int tailLen = LenTmp - sb.length(); 79 while (tailLen > 0) { 80 sb.append(‘ ‘); 81 tailLen--; 82 } 83 84 list.add(sb.toString()); 85 rec(words, LenTmp, end + 1, list); 86 } 87 }
SOLUTION 2:
其实之前的递归是一个尾递归，我们可以很容易地把尾递归转化为Iteration的解法。
以下是Iteration的解法：
思想是一致的。
1 // SOLUTION 2: iteration. 2 public List<String> fullJustify(String[] words, int L) { 3 List<String> ret = new ArrayList<String>(); 4 if (words == null) { 5 return ret; 6 } 7 8 int len = words.length; 9 int index = 0; 10 11 while (index < len) { 12 int LenTmp = L; 13 14 int end = index; 15 for (int i = index; i < len && words[i].length() <= LenTmp; i++) { 16 LenTmp -= words[i].length(); 17 18 // the space follow the word. 19 LenTmp--; 20 end = i; 21 } 22 23 // 最后一个空格收回 24 LenTmp++; 25 26 // Count how many words do we have. 27 int num = end - index + 1; 28 29 int extraSpace = 0; 30 int firstExtra = 0; 31 32 // 单词数大于1，才需要分配，否则所有的空格加到最后即可 33 if (num > 1) { 34 extraSpace = LenTmp / (num - 1); 35 // 首单词后多跟的空格 36 firstExtra = LenTmp % (num - 1); 37 } 38 39 StringBuilder sb = new StringBuilder(); 40 for (int i = index; i <= end; i++) { 41 sb.append(words[i]); 42 43 // The space following every word. 44 if (i != end) { 45 sb.append(‘ ‘); 46 } 47 48 // 不是最后一行 49 if (end != len - 1) { 50 // The first words. 51 if (firstExtra > 0) { 52 sb.append(‘ ‘); 53 firstExtra--; 54 } 55 56 // 最后一个单词后面不需要再加空格 57 if (i == end) { 58 break; 59 } 60 61 // 每个单词后的额外空格 62 int cnt = extraSpace; 63 while (cnt > 0) { 64 sb.append(‘ ‘); 65 cnt--; 66 } 67 } 68 } 69 70 // 最后一行的尾部的空格 71 int tailLen = L - sb.length(); 72 while (tailLen > 0) { 73 sb.append(‘ ‘); 74 tailLen--; 75 } 76 77 ret.add(sb.toString()); 78 index = end + 1; 79 } 80 81 return ret; 82 }
SOLUTION 3:
参考http://www.ninechapter.com/solutions/text-justification/
在solution2的基础上进行了一些简化，把append word的函数独立出来，解答如下，更加简洁：
1 // SOLUTION 3: iteration2 2 public List<String> fullJustify(String[] words, int L) { 3 List<String> ret = new ArrayList<String>(); 4 if (words == null) { 5 return ret; 6 } 7 8 int len = words.length; 9 int index = 0; 10 11 while (index < len) { 12 int LenTmp = L; 13 14 int end = index; 15 for (int i = index; i < len && words[i].length() <= LenTmp; i++) { 16 LenTmp -= words[i].length(); 17 18 // the space follow the word. 19 LenTmp--; 20 end = i; 21 } 22 23 // 最后一个空格收回 24 LenTmp++; 25 26 // Count how many words do we have. 27 int num = end - index + 1; 28 29 int extraSpace = 0; 30 int firstExtra = 0; 31 32 // 单词数大于1，才需要分配，否则所有的空格加到最后即可 33 if (num > 1) { 34 extraSpace = LenTmp / (num - 1); 35 // 首单词后多跟的空格 36 firstExtra = LenTmp % (num - 1); 37 } 38 39 StringBuilder sb = new StringBuilder(); 40 for (int i = index; i <= end; i++) { 41 sb.append(words[i]); 42 43 int cnt = 0; 44 45 if (i == end) { 46 break; 47 } 48 49 // 不是最后一行 50 if (end != len - 1) { 51 // The first words. 52 if (firstExtra > 0) { 53 cnt++; 54 firstExtra--; 55 } 56 57 // 最后一个单词后面不需要再加空格 58 // 每个单词后的额外空格 59 cnt += extraSpace; 60 } 61 62 // 1: 每个单词后本来要加的空格 63 appendSpace(sb, cnt + 1); 64 } 65 66 // 最后一行的尾部的空格，或者是只有一个单词的情况 67 appendSpace(sb, L - sb.length()); 68 69 ret.add(sb.toString()); 70 index = end + 1; 71 } 72 73 return ret; 74 } 75 76 public void appendSpace(StringBuilder sb, int cnt) { 77 while (cnt > 0) { 78 sb.append(‘ ‘); 79 cnt--; 80 } 81 }
GITHUB:
https://github.com/yuzhangcmu/LeetCode_algorithm/blob/master/string/FullJustify.java
LeetCode: Text Justification 解题报告
标签：style blog http io ar color os sp java
原文地址：http://www.cnblogs.com/yuzhangcmu/p/4127290.html
