398. Random Pick Index

类似问题：

382. Linked List Random Node

问题：

给定一组数（可能有重复的数字），

求随机取得数字=target的index。

⚠️ 注意：取得每个数的概率要相同。

Example 1:
Input
["Solution", "pick", "pick", "pick"]
[[[1, 2, 3, 3, 3]], [3], [1], [3]]
Output
[null, 4, 0, 2]

Explanation
Solution solution = new Solution([1, 2, 3, 3, 3]);
solution.pick(3); // It should return either index 2, 3, or 4 randomly. Each index should have equal probability of returning.
solution.pick(1); // It should return 0. Since in the array only nums[0] is equal to 1.
solution.pick(3); // It should return either index 2, 3, or 4 randomly. Each index should have equal probability of returning.
 

Constraints:

1 <= nums.length <= 2 * 104
-231 <= nums[i] <= 231 - 1
target is an integer from nums.
At most 104 calls will be made to pick.

　　

解法：math，概率。

解法一：unordered_map<数字，该数字的所有index(vector)>

初始化构建该map

在随机抽取中：

取得target对应的所有index vector

从中任意取一个：vector[rand()/vector.size]

 

代码参考：

class Solution {
public:
    unordered_map<int, vector<int>> mp;
    Solution(vector<int>& nums) {
        for(int i=0; i<nums.size(); i++) {
            mp[nums[i]].push_back(i);
        }
    }
    
    int pick(int target) {
        int sz = mp[target].size();
        return mp[target][rand()%sz];
    }
};

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

 

解法二：math，概率

思想同382. Linked List Random Node

• 不在初始化做什么。
• 在抽取时，遍历所有数字，
  • 在当前数字的遍历中，若满足概率1/i的时候，进行res赋值。
• 一直到遍历完毕。
• 返回res。

⚠️ 只是，对于本题：当遍历到当前数字，若!=target则continue。不进行任何操作。

只有是target的index，才进行概率计算 i++

⚠️ 注意：在初始化res，为第一个(i==1)访问到的target。

因为此时选取到该index的可能=1/i = 1/1 =1。

 

代码参考：

class Solution {
public:
    vector<int> _num;
    Solution(vector<int>& nums) {
        _num = nums;
    }
    
    int pick(int target) {
        int res = -1;
        int i=0;
        for(int k=0; k<_num.size(); k++) {
            if(_num[k]!=target) continue;
            i++;
            if(i==1) res=k;
            else {
                int rd = rand()%i;
                if(0==rd) res=k;
            }
        }
        return res;
    }
};

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