【Leetcode】Shuffle an Array

题目链接：https://leetcode.com/problems/shuffle-an-array/

题目：

Shuffle a set of numbers without duplicates.

Example:

// Init an array with set 1, 2, and 3.
int[] nums = {1,2,3};
Solution solution = new Solution(nums);// Shuffle the array [1,2,3] and return its result. Any permutation of [1,2,3] must equally likely to be returned.
solution.shuffle();// Resets the array back to its original configuration [1,2,3].
solution.reset();// Returns the random shuffling of array [1,2,3].
solution.shuffle();

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思路：

使用 http://blog.csdn.net/yeqiuzs/article/details/52141124 中实现的数据结构，每次shuffle时先获取初始集合一个随机数并删除，直至集合为空，一次shuffle完成，完成后需要恢复该存储集合的数据结构。时间复杂度shuffle方法 O(n)，reset方法O(n)，构造方法O(n)。

这题我做复杂了。。不要这么做。。

最简单的方法是对数组做n次随机交换。。

算法：

public class Solution {RandomizedSet rm;int nums[];public Solution(int[] nums) {rm = new RandomizedSet();this.nums = nums;for (int i=0;i<nums.length;i++) {rm.insert(nums[i]);}}/** Resets the array to its original configuration and return it. */public int[] reset() {rm = new RandomizedSet();for (int i=0;i<nums.length;i++) {rm.insert(nums[i]);}return nums;}/** Returns a random shuffling of the array. */public int[] shuffle() {RandomizedSet tmp = new RandomizedSet();int[] res = new int[rm.vals.size()];for (int i = 0; i < res.length; i++) {res[i] = rm.getRandom();rm.remove(res[i]);tmp.insert(res[i]);}rm = tmp;return res;}class RandomizedSet {/** Initialize your data structure here. */public RandomizedSet() {}List<Integer> vals = new ArrayList<Integer>();Map<Integer, Integer> val2idx = new HashMap<Integer, Integer>();/*** Inserts a value to the set. Returns true if the set did not already* contain the specified element.*/public boolean insert(int val) {if (val2idx.containsKey(val)) {return false;} else {val2idx.put(val, val2idx.size());vals.add(val);return true;}}/*** Removes a value from the set. Returns true if the set contained the* specified element.*/public boolean remove(int val) {if (!val2idx.containsKey(val)) {return false;} else {int idx = val2idx.remove(val);if (idx < vals.size() - 1) {// 若删除的不是链表最后的元素// 交换末尾元素和被删除元素vals.set(idx, vals.get(vals.size() - 1));val2idx.put(vals.get(vals.size() - 1), idx);}vals.remove(vals.size() - 1);return true;}}/** Get a random element from the set. */Random r = new Random();public int getRandom() {return vals.get(r.nextInt(vals.size()));}}}

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