算法的重要性，我就不多说了吧，想去大厂，就必须要经过基础知识和业务逻辑面试+算法面试。所以，为了提高大家的算法能力，这个公众号后续每天带大家做一道算法题，题目就从LeetCode上面选 ！今天和大家聊的问题叫做 翻转字符串里的单词 II ，我们先来看题面：https:///problems/reverse-words-in-a-string-ii/Given an input string , reverse the string word by word.
Example: Input: ["t","h","e"," ","s","k","y"," ","i","s"," ","b","l","u","e"] Output: ["b","l","u","e"," ","i","s"," ","s","k","y"," ","t","h","e"]
题意示例示例：
输入: ["t","h","e"," ","s","k","y"," ","i","s"," ","b","l","u","e"]
输出: ["b","l","u","e"," ","i","s"," ","s","k","y"," ","t","h","e"]
注意：
单词的定义是不包含空格的一系列字符
输入字符串中不会包含前置或尾随的空格
单词与单词之间永远是以单个空格隔开的
进阶：使用 O(1) 额外空间复杂度的原地解法。 解题
思路大致为：先对每个单词进行翻转，然后对句子进行翻转即可。对单词反转后，单词变得不伦不类，比如 blue 变成了 eulb，并且此时每个单词的相对位置没变。
再对句子进行一次翻转，不仅把每个单词变正常了，且每个单词的相对位置进行了翻转。
翻转的函数实现很简单，给定 str 和待翻转的索引区间即可。class Solution {
public void reverseWords(char[] str) {
int i = 0;
for (int j = 0; j < str.length; j++) { // aTbTc
if (str[j] == ' ') {
reverse(str, i, j);
i = j + 1;
}
}
reverse(str, i, str.length); // 最后一个单词末尾没有空格
System.out.println(String.valueOf(str));
reverse(str, 0, str.length); // 整体再翻转一次
}
/**
* 将 str 的 [i, j] 进行翻转，如 "the" 转换后变成 “eht”
* 注意，[i,j] 是左闭右开
*
* @param str
* @param i
* @param j
*/
private void reverse(char[] str, int i, int j) {
for (int k = i; k < (i + j) / 2; k++) {
char tmp = str[k]; // 位置 k 的元素
int g = j - 1 - k + i; // 位置 k 的对称位置
str[k] = str[g];
str[g] = tmp;
}
}
}好了，今天的文章就到这里，如果觉得有所收获，请顺手点个在看或者转发吧，你们的支持是我最大的动力 。
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