Leetcode 398. Random Pick Index 随机找位置 解题报告

1 解题思想

给了一个数组，包含若干数，可能有重复，然后再给一个数，查找其位置，如果这个数在数组中只出现了一次就返回那个位置，如果不止一次就随机返回一个。

因为不让用额外空间，所以只能暴力查找了。。
关键在于如何做到随机？这个直接看代码，有个模板类似的东西，之前leetcode有一系列的，我还没做，我后面做了细讲，现在就直接看看就好，很简单

2 原题

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);

3 AC解

public class Solution {
    private int[] nums;
    private Random random;

    public Solution(int[] nums) {
        this.nums = nums;
        this.random = new Random();
    }

    public int pick(int target) {
        int result = -1;
        int upbound = 1;
        for (int i = 0; i < nums.length; i++) {
            if (nums[i] == target) {
                /**
                 * 关于随机这里的用法，leetcode之前有很多题，我没来得及系列的做，后面更新
                 * 有模板，先这样
                 * */
                if (random.nextInt(upbound) == 0) {
                    result = i;
                } 
                upbound++;
            }
        }
        return result;
    }
}

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