Leetcode 398 随机数索引

Given an integer array nums 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.

Implement the Solution class:

• Solution(int[] nums) Initializes the object with the array nums.
• int pick(int target) Picks a random index i from nums where nums[i] == target. If there are multiple valid i's, then each index should have an equal probability of returning.

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/random-pick-index

实现方法比较垃圾，差不多是离散化之后记录下每个数字出现的位置，然后记录有几个位置，然后循环输出（竟然真的会检测概率相等）。

class Solution {
public:
    map<int, int> m;
    vector<int> t, now;
    vector<vector<int>> g;
    Solution(vector<int>& nums) {
        vector<int> cop = nums;
        std::sort(nums.begin(), nums.end());
        int N = nums.size(), cnt = 0;
        m.insert(make_pair(nums[0], cnt++));
        nums[0] = m[nums[0]], g.push_back(vector<int>()), t.push_back(0), now.push_back(0);
        for (int i = 1; i < N; i++) {
            if (nums[i] != nums[i-1]) m.insert(make_pair(nums[i], cnt++)), g.push_back(vector<int>()), t.push_back(0), now.push_back(0);
            nums[i] = m[nums[i]];
        }
        for (int i = 0; i < N; i++) {
            g[m[cop[i]]].push_back(i);
            t[m[cop[i]]]++;
        }
    }
    
    int pick(int target) {
        if (m.find(target) == m.end()) return -1;
        else {
            int tem = m[target];
            now[tem] = (now[tem] + 1) % t[tem];
            return g[tem][now[tem]];
        }
    }
};

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

标签：target,nums,int,back,Solution,随机数,push,398,Leetcode
来源： https://www.cnblogs.com/Gensokyo-Alice/p/16192678.html