[抄题]:
Given an input string, reverse the string word by word.
Example:
Input: "the sky is blue
",
Output: "blue is sky the
".
[暴力解法]:
时间分析:
空间分析:
[优化后]:
时间分析:
空间分析:
[奇葩输出条件]:
[奇葩corner case]:
[思维问题]:
打碎成数组后再添加,居然有看不懂的bug。
public class Solution { public String reverseWords(String s) { //ini char[] str = s.toCharArray(); //cc if (str == null || str.length == 0) return s; //3 step3: reverse the whole, word, last reverse(str, 0, str.length - 1); int wordStart = 0; for (int i = 0; i < str.length; i++) { if (str[i] == ' ') { reverse(str, wordStart, i - 1); wordStart = i + 1; } } reverse(str, wordStart, str.length - 1); private void reverse(char[] str, int start, int end) { //do in a while loop while (start < end) { char temp = str[start]; str[start] = str[end]; str[end] = temp; start++; end--; } } //return StringBuilder sb = new StringBuilder(); for (char ch : str) sb.append(ch); return sb.toString(); } }
[英文数据结构或算法,为什么不用别的数据结构或算法]:
[一句话思路]:
用.split("\\s+")把单词分开(特别注意是反斜杠),然后倒贴(单词+空格),最后一个单词不贴空格。
[输入量]:空: 正常情况:特大:特小:程序里处理到的特殊情况:异常情况(不合法不合理的输入):
[画图]:
[一刷]:
反斜杠写成正斜杠就错了
[二刷]:
[三刷]:
[四刷]:
[五刷]:
[五分钟肉眼debug的结果]:
[总结]:
不管输入是字符串还是字符数组,最后一个单词都先不贴,因为后面没有空格
[复杂度]:Time complexity: O(n) Space complexity: O(n)
[算法思想:迭代/递归/分治/贪心]:
[关键模板化代码]:
[其他解法]:
[Follow Up]:
[LC给出的题目变变变]:
[代码风格] :
[是否头一次写此类driver funcion的代码] :
[潜台词] :
public class Solution { public String reverseWords(String s) { //ini String result = new String(); s = s.trim(); //cc if (s == null || s.length() == 0) return result; //split into words String[] words = s.split("\\s+"); //append for n - 1 to 1 for (int i = words.length - 1; i > 0; i--) result += words[i] + " "; //append 0 result += words[0]; //return return result; } }