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);
这道题指明了我们不能用太多的空间，那么省空间的随机方法只有水塘抽样Reservoir Sampling了，LeetCode之前有过两道需要用这种方法的题目Shuffle an Array和Linked List Random Node。那么如果了解了水塘抽样，这道题就不算一道难题了，我们定义两个变量，计数器cnt和返回结果res，我们遍历整个数组，如果数组的值不等于target，直接跳过；如果等于target，计数器加1，然后我们在[0,cnt)范围内随机生成一个数字，如果这个数字是0，我们将res赋值为i即可，参见代码如下：
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;
};
类似题目：
参考资料：
https://discuss.leetcode.com/topic/58371/c-o-n-solution/2
https://discuss.leetcode.com/topic/58297/share-c-o-n-time-solution/2
https://discuss.leetcode.com/topic/58403/bruce-force-java-with-o-n-time-o-1-space
LeetCode All in One 题目讲解汇总(持续更新中…)
转载请注明来源于 Grandyang 的博客 (grandyang.com)，欢迎对文章中的引用来源进行考证，欢迎指出任何有错误或不够清晰的表达。可以在下面评论区评论，也可以邮件至 grandyang@qq.com
