186. Reverse Words in a String II 翻转有空格的单词串 里面不变
[抄题]:
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"]
[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]:
[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]:
全转+空格前单词转+最后一个补转
[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]:
[二刷]:
[三刷]:
[四刷]:
[五刷]:
[五分钟肉眼debug的结果]:
[总结]:
<n时,第n位是不被处理的。需要补充翻转
[复杂度]:Time complexity: O(n) Space complexity: O(1)
[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] :
class Solution { public void reverseWords(char[] str) { //cc if (str == null || str.length == 0) return ; //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); } public 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--; } } }