LeetCode No.398 Random Pick Index

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);

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题目链接：https://leetcode.com/problems/random-pick-index/

题目大意：一个数在数组中可能出现多次，要求等概率返回一个位置。

思路：先用一个数组将所有满足条件的下标记录下来，利用蓄水池抽样原理对它们进行随机抽取。

参考代码：

class Solution {public:    Solution(vector<int> nums) {        this -> nums = nums ;    }        int pick(int target) {        int n = nums.size() ;        vector <int> index ;        for ( int i = 0 ; i < n ; i ++ )            if ( nums[i] == target )                index.push_back ( i ) ;        int sum = index.size() , ans = index[0] ;        for ( int i = 1 ; i < sum ; i ++ )            if ( rand() % ( i + 1 ) == 0 )                ans = index[i] ;        return ans ;    }private:    vector <int> nums ;};/** * Your Solution object will be instantiated and called as such: * Solution obj = new Solution(nums); * int param_1 = obj.pick(target); */