1. 青蛙跳跳FrogJmp Count minimal number of jumps from position X to Y.

青蛙跳跳；

package com.code;public class Test03_1 {public int solution(int X, int Y, int D) {int res = (Y-X)/D+((Y-X)%D==0?0:1);return res;}public static void main(String[] args) {Test03_1 t03 = new Test03_1();System.out.println(t03.solution(10, 85, 30));}
}/**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 = 10Y = 85D = 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
Assume that:X, Y and D are integers within the range [1..1,000,000,000];
X ≤ Y.
Complexity:expected worst-case time complexity is O(1);
expected worst-case space complexity is O(1).* * *
*/

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1. 青蛙跳跳FrogJmp Count minimal number of jumps from position X to Y.

1. 青蛙跳跳FrogJmp Count minimal number of jumps from position X to Y.相关推荐

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