[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--;
}
}
}

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