# 2498. Frog Jump II
###### tags: `Leetcode` `Medium` `Greedy`
Link: https://leetcode.com/problems/frog-jump-ii/description/
## 思路
思路和[这篇](https://leetcode.com/problems/frog-jump-ii/solutions/2897948/java-c-python-max-a-i-a-i-2/)写的一样
由于青蛙先从起点跳到终点 再跳回来 并且路上没有一个点用过两次
所以可以想象成两只青蛙用不同的路径分别从起点往终点跳
那么如何保证每条路径的每两点absolute difference最小的 就是间隔跳
assign stones to two frogs alternatively:

A[0] A[1] A[2] A[3] ...
start frog1 frog2 frog1 ...

If we assign them not alternatively like:

The distance for frog 1 is huge and it's no better than

## Code

class Solution {
    public int maxJump(int[] stones) {
        int max = 0;
        int n = stones.length;
        int prev1 = stones[0], prev2 = stones[0];
        for(int i=1; i<n-1; i++){
            if(i%2==0){
                max = Math.max(max, stones[i]-prev1);
                prev1 = stones[i];
            }
            else{
                max = Math.max(max, stones[i]-prev2);
                prev2 = stones[i];
            }
        }
        max = Math.max(max, Math.max(stones[n-1]-prev1, stones[n-1]-prev2));
        return max;
    }
}