[LeetCode] Random Pick Index 隨機拾取序列

時間 2019-11-07

標籤 leetcode random pick index 隨機拾取序列简体版

原文原文鏈接

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.html

Note:
The array size can be very large. Solution that uses too much extra space will not pass the judge. java

Example: 數組

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);

這道題指明瞭咱們不能用太多的空間，那麼省空間的隨機方法只有水塘抽樣Reservoir Sampling了，LeetCode以前有過兩道須要用這種方法的題目 Shuffle an Array和 Linked List Random Node。那麼若是瞭解了水塘抽樣，這道題就不算一道難題了，咱們定義兩個變量，計數器cnt和返回結果res，咱們遍歷整個數組，若是數組的值不等於target，直接跳過；若是等於target，計數器加1，而後咱們在[0,cnt)範圍內隨機生成一個數字，若是這個數字是0，咱們將res賦值爲i便可，參見代碼以下：

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