leetcode 刷题记录（高频算法面试题汇总）--打乱数组

打乱数组

打乱一个没有重复元素的数组。

示例:

// 以数字集合 1, 2 和 3 初始化数组。
int[] nums = {1,2,3};
Solution solution = new Solution(nums);

// 打乱数组 [1,2,3] 并返回结果。任何 [1,2,3]的排列返回的概率应该相同。
solution.shuffle();

// 重设数组到它的初始状态[1,2,3]。
solution.reset();

// 随机返回数组[1,2,3]打乱后的结果。
solution.shuffle();

[code]class Solution:

def __init__(self, nums: List[int]):
self.origin = nums[:]
self.res = nums

def reset(self) -> List[int]:
"""
Resets the array to its original configuration and return it.
"""
return self.origin

def shuffle(self) -> List[int]:
"""
Returns a random shuffling of the array.
"""
for i in range(len(self.res)):
r = random.randint(i,len(self.res)-1)
self.res[i],self.res[r] = self.res[r],self.res[i]
return self.res

# Your Solution object will be instantiated and called as such:
# obj = Solution(nums)
# param_1 = obj.reset()
# param_2 = obj.shuffle()

[code]class Solution {
private:
vector<int> origin;
public:
Solution(vector<int> nums) {
origin = nums;
}

/** Resets the array to its original configuration and return it. */
vector<int> reset() {
return origin;
}

/** Returns a random shuffling of the array. */
vector<int> shuffle() {
vector<int> res = origin;
int r;
for(int i=0;i<res.size();i++){
r = rand() % (i+1);
if(r!=i)
swap(res[i],res[r]);
}
return res;
}
};

/**
* Your Solution object will be instantiated and called as such:
* Solution obj = new Solution(nums);
* vector<int> param_1 = obj.reset();
* vector<int> param_2 = obj.shuffle();
*/

思路&问题：

1. 先把原始数组保存下来，reset就返回原始数组。shuffle就生成一个[0,nums.size())范围内的随机数，如果不是当前i，就交换次序。循环nums.size()次，就可以打乱数组了
2. python中的随机数函数random.randint(取值范围)