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:

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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);

思路

我们如果以牺牲空间复杂度来换取时间复杂度的话，可以在初始化Solution类的时候指定一个字典对对应的pickup值进行存储。这样就可以在短时间内（接近于O(1)的时间）来实现查找了。这是最原始的方法了。

但是如果又要保证空间复杂度，又要保证时间复杂度，就要采用蓄水池抽样的方法了。详情可以查看参考资料。

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import copy
class Solution(object):
    def __init__(self, nums):
        """
        :type nums: List[int]
        :type numsSize: int
        """
        self.pickmap = {}
        self.nums = nums
    def pick(self, target):
        """
        :type target: int
        :rtype: int
        """
        res = []
        if (self.pickmap.has_key(target)):
            res = self.pickmap[target]
        else:
            for index, i in enumerate(self.nums):
                if (i == target):
                    res.append(index)
            self.pickmap[target] = copy.copy(res)
        return random.choice(res)
# Your Solution object will be instantiated and called as such:
# obj = Solution(nums)
# param_1 = obj.pick(target)

me？https://discuss.leetcode.com/topic/58659/clean-understandable-o-1-momery-o-n-time-java-solution 参考资料：蓄水池抽样——http://www.cnblogs.com/HappyAngel/archive/2011/02/07/1949762.html