Analysis of the third question of the 2020 vivo spring recruit written test
Count mobile phone output on the Nth day
Look at the title description
/**
* 在vivo产线上，每位职工随着对手机加工流程认识的熟悉和经验的增加，日产量也会不断攀升。
* 假设第一天量产1台，接下来2天(即第二、三天)每天量产2件，接下来3天(即第四、五、六天)每天量产3件 ... ...
* 以此类推，请编程计算出第n天总共可以量产的手机数量。
* 输入例子1:
* 11
*
* 输出例子1:
* 35
*
* 例子说明1:
* 第11天工人总共可以量产的手机数量
*/
It is essentially a step function, and the width of the step is related to the number of days. There should be a formulaic solution to this problem, but I'm too vague to think of it.
So give everyone a good understanding solution.
Remember Leetcode's jumping game II? The requirement of that question is to use the least number of hops to jump to the end. We have an O(n) solution to solve that question, that is, to select a jumping point, and when reaching the next jumping point, according to the difference between the previous two jumping points The maximum jump distance selects the new next jump point.
This problem is very similar to the above jumping game, except that the maximum length of each jump is increased by 1, and the solution is given below.
public class Solution {
/**
*
* @param n int整型 第n天
* @return int整型
*/
public int solution (int n) {
int currGap=1;
int nextPos=1;//需要更新每天生产数量变化的地方
int phoneNum=0;
for(int i=1;i<=n;i++){
phoneNum+=currGap;
if(i==nextPos){
currGap++;
nextPos=i+currGap;
}
}
return phoneNum;
}
}