Given an array of integers with possible duplicates, randomly output the index of a given target number. You can assume that the given target number must exist in the array.
Note: The array size can be very large. Solution that uses too much extra space will not pass the judge.
Example:
int[] nums = new int[] {1,2,3,3,3}; Solution solution = new Solution(nums); // pick(3) should return either index 2, 3, or 4 randomly. Each index should have equal probability of returning. solution.pick(3); // pick(1) should return 0. Since in the array only nums[0] is equal to 1. solution.pick(1);
这道题是在等于特定值的元素中随机取一个，题目难度为Medium。
由于题目限定数组会比较大，而且空间复杂度不能太高，所有我们不能先在Hash Table中存每个元素出现的位置然后随机取一个，这里需要用到蓄水池抽样。蓄水池抽样的算法原理及证明见第382题（传送门），这里不再详述。
这道题中我们只需遍历数组，统计target到当前位置出现的次数，然后依据当前计数进行蓄水池抽样即可。具体代码：
class Solution { vector<int> nums_; public: Solution(vector<int> nums) : nums_(nums){} int pick(int target) { int cnt = 0, sz = nums_.size(), idx = 0; for(int i=0; i<sz; ++i) { if(nums_[i] == target) { if(cnt == 0) idx = i; else { if(rand() % (cnt+1) == 0) idx = i; } ++cnt; } } return idx; } }; /** * Your Solution object will be instantiated and called as such: * Solution obj = new Solution(nums); * int param_1 = obj.pick(target); */