LeetCode 398. Random Pick Index（C++）_leetcode(c++) 柳婼-优快云博客

本文链接：https://blog.youkuaiyun.com/liuchuo/article/details/54670519

本文介绍了一种在含重复元素的整数数组中随机选择指定目标值索引的方法。通过遍历数组找到所有匹配项的位置，并使用随机数生成器均匀地从中挑选一个返回。

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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);

题目大意：给一个整数数组，里面可能有重复元素。给一个target，随机返回一个数组元素等于target的元素下标～
分析：将所有等于target的元素的下标存储在index数组中，然后根据rand() % index.size()的值随机从index中取一个值返回～

class Solution {
private:
    vector<int> nums;
public:
    Solution(vector<int> nums) {
        this->nums = nums;
    }
    
    int pick(int target) {
        vector<int> index;
        for(int i = 0; i < nums.size(); i++) {
            if(nums[i] == target)
                index.push_back(i);
        }
        return index[rand() % index.size()];
    }
};

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