【LeetCode】 384. 打乱数组

Myth丶恋晨 2021-06-10 20:38 476阅读 0赞

## <table>   <tbody>    <tr>     <td><font>题目</font></td>    </tr>   </tbody>  </table> ##

[题目传送门：传送门（点击此处）][Link 1]  
![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxMTUxNTMxMjgzMg_size_16_color_FFFFFF_t_70]

## <table>   <tbody>    <tr>     <td><font>题解</font></td>    </tr>   </tbody>  </table> ##

1.  直接看到题目，也应该想到要用随机数了，第一点要考虑的就是随机数怎么用
2.  考虑到有一个reset()，还有一个shuffle()，很简单的办法，克隆一份保存起来，因为数组是传引用的，所以不可以直接赋值
3.  我在考虑了随机数之后，思路其实还蛮简单的，根据索引，在数组长度范围内，随机取一个下标的值和当前索引的值交换

### 代码 ###

class Solution { 
    
        int[] nums;
        int[] backNums; // 数组备份
    
        public Solution(int[] nums) { 
            this.nums = nums;
            backNums = nums.clone();
        }
    
        /** * Resets the array to its original configuration and return it. */
        public int[] reset() { 
            nums = backNums.clone();
            return nums;
        }
    
        /** * Returns a random shuffling of the array. */
        public int[] shuffle() { 
            int n = nums.length;
            Random random = new Random();
            for (int i = 0; i < n; i++) { 
                int p = random.nextInt(n);
                int temp = nums[i];
                nums[i] = nums[p];
                nums[p] = temp;
            }
            return nums;
        }
    }
    
    /** * Your Solution object will be instantiated and called as such: * Solution obj = new Solution(nums); * int[] param_1 = obj.reset(); * int[] param_2 = obj.shuffle(); */

[Link 1]: https://leetcode-cn.com/problems/shuffle-an-array/
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