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1 | 1 | package com.fishercoder.solutions;
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2 | 2 |
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3 |
| -/** |
4 |
| - * 186. Reverse Words in a String II |
5 |
| -
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6 |
| - Given an input string, reverse the string word by word. A word is defined as a sequence of non-space characters. |
7 |
| -
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8 |
| - The input string does not contain leading or trailing spaces and the words are always separated by a single space. |
9 |
| -
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10 |
| - For example, |
11 |
| - Given s = "the sky is blue", |
12 |
| - return "blue is sky the". |
13 |
| -
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14 |
| - Could you do it in-place without allocating extra space? |
15 |
| - */ |
16 | 3 | public class _186 {
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17 |
| - public static class Solution1 { |
18 |
| - public void reverseWords(char[] s) { |
19 |
| - // Three steps to reverse |
20 |
| - // 1, reverse the whole sentence |
21 |
| - reverse(s, 0, s.length - 1); |
22 |
| - // 2, reverse each word |
23 |
| - int start = 0; |
24 |
| - for (int i = 0; i < s.length; i++) { |
25 |
| - if (s[i] == ' ') { |
26 |
| - reverse(s, start, i - 1); |
27 |
| - start = i + 1; |
| 4 | + public static class Solution1 { |
| 5 | + public void reverseWords(char[] s) { |
| 6 | + // Three steps to reverse |
| 7 | + // 1, reverse the whole sentence |
| 8 | + reverse(s, 0, s.length - 1); |
| 9 | + // 2, reverse each word |
| 10 | + int start = 0; |
| 11 | + for (int i = 0; i < s.length; i++) { |
| 12 | + if (s[i] == ' ') { |
| 13 | + reverse(s, start, i - 1); |
| 14 | + start = i + 1; |
| 15 | + } |
| 16 | + } |
| 17 | + // 3, reverse the last word, if there is only one word this will solve the corner case |
| 18 | + reverse(s, start, s.length - 1); |
| 19 | + } |
| 20 | + |
| 21 | + private void reverse(char[] s, int start, int end) { |
| 22 | + while (start < end) { |
| 23 | + char temp = s[start]; |
| 24 | + s[start++] = s[end]; |
| 25 | + s[end--] = temp; |
| 26 | + } |
28 | 27 | }
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29 |
| - } |
30 |
| - // 3, reverse the last word, if there is only one word this will solve the corner case |
31 |
| - reverse(s, start, s.length - 1); |
32 | 28 | }
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33 | 29 |
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34 |
| - private void reverse(char[] s, int start, int end) { |
35 |
| - while (start < end) { |
36 |
| - char temp = s[start]; |
37 |
| - s[start++] = s[end]; |
38 |
| - s[end--] = temp; |
39 |
| - } |
| 30 | + public static class Solution2 { |
| 31 | + public void reverseWords(char[] s) { |
| 32 | + reverse(s, 0, s.length); |
| 33 | + for (int i = 0; i < s.length; i++) { |
| 34 | + int start = i; |
| 35 | + while (i < s.length && s[i] != ' ') { |
| 36 | + i++; |
| 37 | + } |
| 38 | + reverse(s, start, i); |
| 39 | + } |
| 40 | + } |
| 41 | + |
| 42 | + private void reverse(char[] chars, int start, int end) { |
| 43 | + int left = start; |
| 44 | + int right = end - 1; |
| 45 | + while (left < right) { |
| 46 | + char tmp = chars[left]; |
| 47 | + chars[left] = chars[right]; |
| 48 | + chars[right] = tmp; |
| 49 | + left++; |
| 50 | + right--; |
| 51 | + } |
| 52 | + } |
40 | 53 | }
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41 |
| - } |
42 | 54 |
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43 | 55 | }
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