feat: add 068

Author: Blankj

README.md

@@ -91,6 +91,7 @@
 |25|[Reverse Nodes in k-Group][025]|Linked List|
 |44|[Reverse Nodes in k-Group][044]|String, Dynamic Programming, Backtracking, Greedy|
 |57|[Insert Interval][057]|Array, Sort|
+|68|[Text Justification][068]|String|
 
 
 
@@ -152,3 +153,4 @@
 [025]: https://github.com/Blankj/awesome-java-leetcode/blob/master/note/025/README.md
 [044]: https://github.com/Blankj/awesome-java-leetcode/blob/master/note/044/README.md
 [057]: https://github.com/Blankj/awesome-java-leetcode/blob/master/note/057/README.md
+[068]: https://github.com/Blankj/awesome-java-leetcode/blob/master/note/068/README.md

note/068/README.md

@@ -0,0 +1,95 @@
+# [Text Justification][title]
+
+## Description
+
+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.
+
+**Corner Cases:**
+
+- 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.
+
+**Tags:** String
+
+
+## 思路
+
+题意是给你一组单词和最大行宽，让你对齐他们，对齐的规则就是尽可能一行可以放下足够多的单词，如果最后有多余的空格，那就把空格均匀地插入到单词之间，如果不能平分的话，那就从左开始依次多插一个空格，最后一行单词之间就正常地一个空格即可，如果凑不到最大行宽，那就在末尾补充空格即可，描述地比较差，不懂的话其实看看demo也就懂了哈。题还是比较坑的，毕竟踩的比赞的人多，我也是靠模拟老老实实做出来的，求出可以最多插入空格数，然后用它除以可以插入的槽数获取每个单词之间的空格，它两取余的话就是最面需要多插入一个空格的个数，最后一行的话就单独处理即可。
+
+```java
+class Solution {
+ public List<String> fullJustify(String[] words, int maxWidth) {
+ int len = words.length;
+ List<String> ans = new ArrayList<>();
+ StringBuilder spaces = new StringBuilder();
+ for (int i = 0; i < maxWidth; ++i) {
+ spaces.append(" ");
+ }
+ int curLen = -1, start = 0;
+ for (int i = 0; i < len; ++i) {
+ if (curLen + words[i].length() + 1 <= maxWidth) {
+ curLen += words[i].length() + 1;
+ } else {
+ StringBuilder sub = new StringBuilder(words[start]);
+ int rest = maxWidth - curLen;
+ int l = i - start - 1;
+ if (l <= 0) {
+ sub.append(spaces.substring(0, rest));
+ } else {
+ int m = rest / l + 1;
+ int mod = rest % l;
+ for (int j = start + 1; j < i; ++j) {
+ if (mod-- > 0) {
+ sub.append(spaces.substring(0, m + 1)).append(words[j]);
+ } else {
+ sub.append(spaces.substring(0, m)).append(words[j]);
+ }
+ }
+ }
+ ans.add(sub.toString());
+ start = i;
+ curLen = words[i].length();
+ }
+ }
+ StringBuilder sub = new StringBuilder(words[start]);
+ for (int i = start + 1; i < len; ++i) {
+ sub.append(" ").append(words[i]);
+ }
+ ans.add(sub + spaces.substring(0, maxWidth - sub.length()));
+ return ans;
+ }
+}
+```
+
+
+## 结语
+
+如果你同我一样热爱数据结构、算法、LeetCode，可以关注我GitHub上的LeetCode题解：[awesome-java-leetcode][ajl]
+
+
+
+[title]: https://leetcode.com/problems/text-justification
+[ajl]: https://github.com/Blankj/awesome-java-leetcode

src/com/blankj/hard/_068/Solution.java

@@ -0,0 +1,61 @@
+package com.blankj.hard._068;
+
+import java.util.ArrayList;
+import java.util.List;
+
+/**
+ * <pre>
+ * author: Blankj
+ * blog : http://blankj.com
+ * time : 2017/11/01
+ * desc :
+ * </pre>
+ */
+public class Solution {
+
+ public List<String> fullJustify(String[] words, int maxWidth) {
+ int len = words.length;
+ List<String> ans = new ArrayList<>();
+ StringBuilder spaces = new StringBuilder();
+ for (int i = 0; i < maxWidth; ++i) {
+ spaces.append(" ");
+ }
+ int curLen = -1, start = 0;
+ for (int i = 0; i < len; ++i) {
+ if (curLen + words[i].length() + 1 <= maxWidth) {
+ curLen += words[i].length() + 1;
+ } else {
+ StringBuilder sub = new StringBuilder(words[start]);
+ int rest = maxWidth - curLen;
+ int l = i - start - 1;
+ if (l <= 0) {
+ sub.append(spaces.substring(0, rest));
+ } else {
+ int m = rest / l + 1;
+ int mod = rest % l;
+ for (int j = start + 1; j < i; ++j) {
+ if (mod-- > 0) {
+ sub.append(spaces.substring(0, m + 1)).append(words[j]);
+ } else {
+ sub.append(spaces.substring(0, m)).append(words[j]);
+ }
+ }
+ }
+ ans.add(sub.toString());
+ start = i;
+ curLen = words[i].length();
+ }
+ }
+ StringBuilder sub = new StringBuilder(words[start]);
+ for (int i = start + 1; i < len; ++i) {
+ sub.append(" ").append(words[i]);
+ }
+ ans.add(sub + spaces.substring(0, maxWidth - sub.length()));
+ return ans;
+ }
+
+ public static void main(String[] args) {
+ Solution solution = new Solution();
+ System.out.println(solution.fullJustify(new String[]{"This", "is", "an", "example", "of", "text", "justification."}, 16));
+ }
+}