leetcode-68. Text Justification

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leetcode-68. Text Justification

题目：

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.

从思路上说这个代码主要分为3个部分。
第一部分判断当前行可以容纳多少个词组
第二部分判断插入多少个空格剩余多少个空格
第三部分加入list结果中。
答案来源

public class Solution {public List<String> fullJustify(String[] words, int L) {List<String> list = new LinkedList<String>();for (int i = 0, w; i < words.length; i = w) {int len = -1;for (w = i; w < words.length && len + words[w].length() + 1 <= L; w++) {len += words[w].length() + 1;}StringBuilder strBuilder = new StringBuilder(words[i]);int space = 1, extra = 0;if (w != i + 1 && w != words.length) { // not 1 char, not last linespace = (L - len) / (w - i - 1) + 1;extra = (L - len) % (w - i - 1);}for (int j = i + 1; j < w; j++) {for (int s = space; s > 0; s--) strBuilder.append(' ');if (extra-- > 0) strBuilder.append(' ');strBuilder.append(words[j]);}int strLen = L - strBuilder.length();while (strLen-- > 0) strBuilder.append(' ');list.add(strBuilder.toString());}return list;}
}