[Leetcode] 398. Random Pick Index 解题报告-优快云博客

本文链接：https://blog.youkuaiyun.com/magicbean2/article/details/77865765

介绍了一种用于从含重复元素的大规模整数数组中随机选取目标数字索引的高效算法。该算法采用水池采样策略，仅需O(n)时间复杂度与O(1)的空间复杂度即可完成任务。

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题目：

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);

思路：

对于大数据而言水池采样是神器：我们遍历数组的时候，如果发现了新的target，则让它以一定的概率转移到该索引上。有兴趣的读者可以参考水池采样的问题描述及正确性证明：https://en.wikipedia.org/wiki/Reservoir_sampling。该算法的时间复杂度是O(n)，其中n是数组的长度；空间复杂度是O(1)。

代码：

class Solution {
public:
    Solution(vector<int> nums) {
        vec = nums;
    }
    
    int pick(int target) {
        int ans = 0, cnt = 1;
        for(int i = 0; i < vec.size(); ++i) {  
            if(vec[i] == target && rand() % (cnt++) == 0) {
                ans = i;
            }
        }
        return ans;  
    }
private:
    vector<int> vec;
};

/**
 * Your Solution object will be instantiated and called as such:
 * Solution obj = new Solution(nums);
 * int param_1 = obj.pick(target);
 */