Random Pick Index
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);
class Solution {
int[] nums;
Random rand;
public Solution(int[] nums) {
this.nums = nums;
this.rand = new Random();
}
public int pick(int target) {
int count = 0;
int sample = -1;
for(int i = 0; i < nums.length; i++){
if(nums[i] == target){
count++;
int x = rand.nextInt(count);
if(x == 0){
sample = i;
}
}
}
return sample;
}
}
a b c c c
c
sample = 2 3 ½ 3 ⅓
2 ½ 2 ⅓
4 ⅓
counter = 3; [0, 3)
i = 3
index = 2, 3
counter = 1 2
sample = 2
0, 1, 2
sample 0 0 ½ 0 ½ * ⅔ = ⅓
1 ½
2 ⅓
counter 1, 2, 3
产生一个随机数,如果这个随机数等于0, 替换掉原来的index, 不然保留
sample = 0
counter = 1
for (int i = 1; i < … ;i++) {
counter++;
int random = randint(0, counter);
if (random == 0) { 1/counter 替换掉原来的数字
sample = i;
}
}
P(第三步保留的是0) = P(第二步是0
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);
来源:https://www.cnblogs.com/tobeabetterpig/p/9550487.html