Leetcode刷题记代码优化记录

﹏ヽ暗。殇╰゛Y 2021-06-11 15:14 373阅读 0赞

目前刚看完数组的基础篇，再看数组的基础题目，这次的题目是832题，属于简单类型  
[https://leetcode.com/problems/flipping-an-image/][https_leetcode.com_problems_flipping-an-image]

# Flipping an Image #

> Given a binary matrix A, we want to flip the image horizontally, then invert it, and return the resulting image.  
>   
>   
> To flip an image horizontally means that each row of the image is reversed. For example, flipping \[1, 1, 0\] horizontally results in \[0, 1, 1\].  
>   
>   
> To invert an image means that each 0 is replaced by 1, and each 1 is replaced by 0. For example, inverting \[0, 1, 1\] results in \[1, 0, 0\].

## Example 1: ##

> **Input:** \[\[1,1,0\],\[1,0,1\],\[0,0,0\]\]  
> **Output:** \[\[1,0,0\],\[0,1,0\],\[1,1,1\]\]  
> **Explanation:** First reverse each row: \[\[0,1,1\],\[1,0,1\],\[0,0,0\]\].  
> Then, invert the image: \[\[1,0,0\],\[0,1,0\],\[1,1,1\]\]

## Example 2: ##

> **Input:** \[\[1,1,0,0\],\[1,0,0,1\],\[0,1,1,1\],\[1,0,1,0\]\]  
> **Output:** \[\[1,1,0,0\],\[0,1,1,0\],\[0,0,0,1\],\[1,0,1,0\]\]  
> **Explanation:** First reverse each row: \[\[0,0,1,1\],\[1,0,0,1\],\[1,1,1,0\],\[0,1,0,1\]\].  
> Then invert the image: \[\[1,1,0,0\],\[0,1,1,0\],\[0,0,0,1\],\[1,0,1,0\]\]

## Notes: ##

> 1 <= A.length = A\[0\].length <= 20  
> 0 <= A\[i\]\[j\] <= 1

# 第一次代码提交 #

第一次我重新定义了一个新的二维数组，用来存放，所以就会特别的麻烦。

//1. 第一版
      
        public static int[][] flipAndInvertImage(int[][] A) { 
            int m = A.length;
            if (m == 0) return A;
            int n = A[0].length == 0 ? 0 : A[0].length;
            int[][] B = new int[m][n];
            for (int i = 0; i < m; i++) { 
                for (int j = 0; j < n; j++) { 
                    B[i][j] = A[i][j];
                }
            }
            for (int i = 0; i < m; i++) { 
                for (int j = 0; j < n; j++) { 
                    B[i][A[i].length - 1 - j] = A[i][j] == 1 ? 0 : 1;
                }
            }
            return B;
        }

# 第二版 #

这一版进行了更新，我直接在数组A上面进行了操作，这样不用重新定义数组B，不需要开辟数组A那么大得一块儿空间

//2.第二版
    
        public static int[][] flipAndInvertImage(int[][] A) { 
            int m = A.length;
            if (m == 0) return A;
            int n = A[0].length == 0 ? 0 : A[0].length - 1;
            int temp;
            for (int i = 0; i < m; i++) { 
                for (int j = 0; j < n / 2 + 1; j++) { 
                    temp = A[i][j];
                    A[i][j] = A[i][n - j] == 0 ? 1 : 0;
                    A[i][n - j] = temp == 0 ? 1 : 0;
                }
            }
            return A;
        }

# 第三版 #

最后一版，我又画了for循环，这样看起来更加清晰，操作起来更加的方便。

// 3. 第三版
        
        public static int[][] flipAndInvertImage(int[][] A) { 
            int m = A.length;
            if (m == 0) return A;
            int temp, n;
            for (int nums[] : A) { 
                n = nums.length - 1;
                for (int j = 0; 2 * j <= n; j++) { 
                    temp = nums[j];
                    nums[j] = nums[n - j] == 0 ? 1 : 0;
                    nums[n - j] = temp == 0 ? 1 : 0;
                }
            }
            return A;
        }

# 总结 #

三次提交代码的性能都有变化，可以从这其中发现很多可以优化的地方，也很有意思  
下面这是三次散发性能的结果  
![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxMTUxNTMxMjgzMg_size_16_color_FFFFFF_t_70]

# 不同语言对比 #

有大神给出了三种语言的代码，完全体现出了py代码的简洁，还体现出了Java代码巧妙地利用方式  
![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxMTUxNTMxMjgzMg_size_16_color_FFFFFF_t_70 1]

[https_leetcode.com_problems_flipping-an-image]: https://leetcode.com/problems/flipping-an-image/
[watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxMTUxNTMxMjgzMg_size_16_color_FFFFFF_t_70]: /images/20210529/19a12931681f4a44bcdb4aaf66bdd5b4.png
[watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxMTUxNTMxMjgzMg_size_16_color_FFFFFF_t_70 1]: /images/20210529/23c518ad66cf4ef681a52fb33aa3b40c.png