You are given an integer array A.  From some starting index, you can make a series of jumps.  The (1st, 3rd, 5th, ...) jumps in the series are called odd numbered jumps, and the (2nd, 4th, 6th, ...) jumps in the series are called even numbered jumps.

You may from index i

• During odd numbered jumps (ie. jumps 1, 3, 5, ...), you jump to the index j such that A[i] <= A[j] and A[j] is the smallest possible value.  If there are multiple such indexes j
  , you can only jump to the smallest such index j.
• During even numbered jumps (ie. jumps 2, 4, 6, ...), you jump to the index j such that A[i] >= A[j] and A[j] is the largest possible value.  If there are multiple such indexes j, you can only jump to the smallest
   such index j.
• (It may be the case that for some index i, there are no legal jumps.)

A starting index is good if, starting from that index, you can reach the end of the array (index A.length - 1) by jumping some number of times (possibly 0 or more than once.)

Return the number of good starting indexes.

Example 1:

Input: [10,13,12,14,15]
Output: 2
Explanation: 
From starting index i = 0, we can jump to i = 2 (since A[2] is the smallest among A[1], A[2], A[3], A[4] that is greater or equal to A[0]), then we can't jump any more.
From starting index i = 1 and i = 2, we can jump to i = 3, then we can't jump any more.
From starting index i = 3, we can jump to i = 4, so we've reached the end.
From starting index i = 4, we've reached the end already.
In total, there are 2 different starting indexes (i = 3, i = 4) where we can reach the end with some number of jumps.

Example 2:

Input: [2,3,1,1,4]
Output: 3
Explanation: 
From starting index i = 0, we make jumps to i = 1, i = 2, i = 3:

During our 1st jump (odd numbered), we first jump to i = 1 because A[1] is the smallest value in (A[1], A[2], A[3], A[4]) that is greater than or equal to A[0].

During our 2nd jump (even numbered), we jump from i = 1 to i = 2 because A[2] is the largest value in (A[2], A[3], A[4]) that is less than or equal to A[1].  A[3] is also the largest value, but 2 is a smaller index, so we can only jump to i = 2 and not i = 3.

During our 3rd jump (odd numbered), we jump from i = 2 to i = 3 because A[3] is the smallest value in (A[3], A[4]) that is greater than or equal to A[2].

We can't jump from i = 3 to i = 4, so the starting index i = 0 is not good.

In a similar manner, we can deduce that:
From starting index i = 1, we jump to i = 4, so we reach the end.
From starting index i = 2, we jump to i = 3, and then we can't jump anymore.
From starting index i = 3, we jump to i = 4, so we reach the end.
From starting index i = 4, we are already at the end.
In total, there are 3 different starting indexes (i = 1, i = 3, i = 4) where we can reach the end with some number of jumps.

Example 3:

Input: [5,1,3,4,2]
Output: 3
Explanation: 
We can reach the end from starting indexes 1, 2, and 4.

Note:

1. 1 <= A.length <= 20000
2. 0 <= A[i] < 100000

思路：關鍵是要計算出以下：

1. 對於index為i的位置，後面比A[i]大的最小數對應的index

2. 對於index為i的位置，後面比A[i]小的最大數對應的index

然後就可以從後往前DP了，

1. dp[i][0]表示到第index為i，並且下一步是奇數步的情況下，能不能到最後

2. dp[i][1]表示到第index為i，並且下一步是偶數步的情況下，能不能到最後

狀態轉移方程就是要用到上面預先計算出來的陣列

PS：Java直接TreeMap就搞定了，Python應該是要維護單調陣列然後bisect二分來找，太麻煩了。。。。

import java.util.HashMap;
import java.util.Map;
import java.util.TreeMap;

class Solution {
    public int oddEvenJumps(int[] A) {
    	Map<Integer, Integer>larger_smallest = new HashMap<Integer, Integer>();
    	Map<Integer, Integer>small_largest = new HashMap<Integer, Integer>();
        TreeMap<Integer, Integer>helper = new TreeMap<Integer, Integer>();
        for(int i=A.length-1; i>=0; i--) {
        	larger_smallest.put(i, -1); 
        	small_largest.put(i, -1);
        	
        	Integer t1 = helper.ceilingKey(A[i]);
        	Integer t2 = helper.floorKey(A[i]);
        	if(t1!=null) larger_smallest.put(i, helper.get(t1));
        	if(t2!=null) small_largest.put(i, helper.get(t2));
        	
        	helper.put(A[i], i);
        }
        
        
        boolean[][] dp = new boolean[A.length][2];
        dp[A.length-1][0] = true;
        dp[A.length-1][1] = true;
        for(int i=A.length-2; i>=0; i--) {
        	if(larger_smallest.get(i)!=-1) dp[i][0]=dp[larger_smallest.get(i)][1];
        	if(small_largest.get(i)!=-1)   dp[i][1]=dp[small_largest.get(i)][0];
        }
        
        int res=0;
        for(int i=0;i<dp.length;i++){
        	if(dp[i][0])res++;
        }
        return res;
    }
}