Time Complexity - FrogJmp

Question

A small frog wants to get to the other side of the road. The frog is currently located at position X and wants to get to a position greater than or equal to Y. The small frog always jumps a fixed distance, D.

Count the minimal number of jumps that the small frog must perform to reach its target.

Write a function:

class Solution { public int solution(int X, int Y, int D); }

that, given three integers X, Y and D, returns the minimal number of jumps from position X to a position equal to or greater than Y.

For example, given:

X = 10 Y = 85 D = 30

the function should return 3, because the frog will be positioned as follows:

• after the first jump, at position 10 + 30 = 40
• after the second jump, at position 10 + 30 + 30 = 70
• after the third jump, at position 10 + 30 + 30 + 30 = 100

Write an efficient algorithm for the following assumptions:

• X, Y and D are integers within the range [1..1,000,000,000];
• X ≤ Y.

要注意题目里写的efficient，report里会考虑到算法复杂度。我一开始用累加就判定为非常复杂了，于是改用取模。

MyReport

class Solution {
    public int solution(int X, int Y, int D) {
        // write your code in Java SE 8
        int distant = Y - X;
        int step = 0;
        if(distant % D == 0) {
            return distant/D;
        } else {
            return distant/D + 1;
        }
    }
}