Open serendipityApe opened 2 years ago
题目
重置一个数组,(洗牌) 原题目: can you shuffle() an array?
例子
如let arr = [1,2,3,4] 可能有4 3 2 * 1 24种排列。 [1, 2, 3, 4] [1, 2, 4, 3] [1, 3, 2, 4] [1, 3, 4, 2] [1, 4, 2, 3] [1, 4, 3, 2] [2, 1, 3, 4] [2, 1, 4, 3] [2, 3, 1, 4] [2, 3, 4, 1] [2, 4, 1, 3] [2, 4, 3, 1] [3, 1, 2, 4] [3, 1, 4, 2] [3, 2, 1, 4] [3, 2, 4, 1] [3, 4, 1, 2] [3, 4, 2, 1] [4, 1, 2, 3] [4, 1, 3, 2] [4, 2, 1, 3] [4, 2, 3, 1] [4, 3, 1, 2] [4, 3, 2, 1]
let arr = [1,2,3,4]
[1, 2, 3, 4] [1, 2, 4, 3] [1, 3, 2, 4] [1, 3, 4, 2] [1, 4, 2, 3] [1, 4, 3, 2] [2, 1, 3, 4] [2, 1, 4, 3] [2, 3, 1, 4] [2, 3, 4, 1] [2, 4, 1, 3] [2, 4, 3, 1] [3, 1, 2, 4] [3, 1, 4, 2] [3, 2, 1, 4] [3, 2, 4, 1] [3, 4, 1, 2] [3, 4, 2, 1] [4, 1, 2, 3] [4, 1, 3, 2] [4, 2, 1, 3] [4, 2, 3, 1] [4, 3, 1, 2] [4, 3, 2, 1]
shuffle方法应该以相同的概率(1/24)讲arr转换成上述任意一个数组。
答案
// 数组第一个元素被置换的几率为1/n; // 第二个元素被置换的几率为 n-1/n * 1/n-1 = 1/n // 以此类推,每个元素被置换的几率都为 1/n function shuffle(arr){ for(let i=0;i<arr.length-1;i++){ let j = i + parseInt(Math.random() *(arr.length - i)); [ arr[i] , arr[j] ] = [ arr[j] , arr[i] ]; } }
重置一个数组,(洗牌) 原题目: can you shuffle() an array?
如
let arr = [1,2,3,4]
可能有4 3 2 * 1 24种排列。[1, 2, 3, 4] [1, 2, 4, 3] [1, 3, 2, 4] [1, 3, 4, 2] [1, 4, 2, 3] [1, 4, 3, 2] [2, 1, 3, 4] [2, 1, 4, 3] [2, 3, 1, 4] [2, 3, 4, 1] [2, 4, 1, 3] [2, 4, 3, 1] [3, 1, 2, 4] [3, 1, 4, 2] [3, 2, 1, 4] [3, 2, 4, 1] [3, 4, 1, 2] [3, 4, 2, 1] [4, 1, 2, 3] [4, 1, 3, 2] [4, 2, 1, 3] [4, 2, 3, 1] [4, 3, 1, 2] [4, 3, 2, 1]
shuffle方法应该以相同的概率(1/24)讲arr转换成上述任意一个数组。