48 lines
1.3 KiB
Java
48 lines
1.3 KiB
Java
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# 58.1 翻转单词顺序列
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[NowCoder](https://www.nowcoder.com/practice/3194a4f4cf814f63919d0790578d51f3?tpId=13&tqId=11197&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking&from=cyc_github)
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## 题目描述
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```html
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Input:
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"I am a student."
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Output:
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"student. a am I"
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```
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## 解题思路
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题目应该有一个隐含条件，就是不能用额外的空间。虽然 Java 的题目输入参数为 String 类型，需要先创建一个字符数组使得空间复杂度为 O(N)，但是正确的参数类型应该和原书一样，为字符数组，并且只能使用该字符数组的空间。任何使用了额外空间的解法在面试时都会大打折扣，包括递归解法。
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正确的解法应该是和书上一样，先旋转每个单词，再旋转整个字符串。
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```java
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public String ReverseSentence(String str) {
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int n = str.length();
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char[] chars = str.toCharArray();
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int i = 0, j = 0;
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while (j <= n) {
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if (j == n || chars[j] == ' ') {
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reverse(chars, i, j - 1);
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i = j + 1;
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}
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j++;
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}
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reverse(chars, 0, n - 1);
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return new String(chars);
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}
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private void reverse(char[] c, int i, int j) {
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while (i < j)
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swap(c, i++, j--);
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}
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private void swap(char[] c, int i, int j) {
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char t = c[i];
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c[i] = c[j];
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c[j] = t;
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}
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```
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