186. Reverse Words in a String II 翻转有空格的单词串 里面不变

时间:2021-07-23 14:31:53

[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] :

 

186. Reverse Words in a String II 翻转有空格的单词串 里面不变186. Reverse Words in a String II 翻转有空格的单词串 里面不变
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--;
        }
    }
}
View Code