题目描述：

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 jump forward to index j (with i < j) in the following way:

  • 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

解析：

从任意的start idx开始跳，进行连续的条约，分为1/3/5奇数跳或者2/4/6偶数跳，但是有如下的规则：

1.跳数为奇数的情况下，跳到后续位置中一个最小值处，但是这个最小值A[j]>=A[i]是要成立的

2.跳数为偶数的情况下，跳到后续位置中一个最大值处，但是这个最大值A[j]<=A[i]是成立的

3.可能存在某些起始点处，无法存在合法的跳法

最终要返回start idx的数量（从这些点开始能跳到数组末尾） 

/*
考虑从后往前进行，以[5,1,3,4,2]为例，根据题目描述的，我们每次应该根据次数找lower或者higher值，并且标记是否能够达到末尾。
（1）对于2来说，lower(2)=true,higher(2)=true，因为已经在末尾
（2）对于4来说，lower(4)=true,higher(4)=false
（3）对于3来说，lower(3)=true,higher(3)=lower(4)=true
（4）对于1来说，lower(1)=false,higher(1)=lower(2)=true
（5）对于5来说，lower(5)=higher(4)=false,higher(5)=false
*/
class Solution {
public:
    int oddEvenJumps(vector<int>& A) {
        int n=A.size();
        vector<int> lower(n);
        vector<int> higher(n); //0标识不可达，1标识可达
        lower[n-1]=higher[n-1]=1; //末尾位置可达
        mymap[A[n-1]]=n-1; //初始化最后一个点的坐标
        for(int i=n-2;i>=0;--i){
            //对于当前的A[i]，我需要在后续中分别找到大于及小于它的位置
            auto lo=mymap.upper_bound(A[i]); //后续中第一个大于，但我们目标是小于，因此后续先--lo
            auto hi=mymap.lower_bound(A[i]);  //后续中第一个大于等于
            if(hi!=mymap.end())
                higher[i]=lower[hi->second]; 
            if(lo!=mymap.begin()) //因为我们迭代器需要先--，因此判断是否是首位置
                lower[i]=higher[(--lo)->second];
            if(higher[i]) //以higher来check即可
                res++;
            mymap[A[i]]=i; //记录当前序号
        }
        return res;
    }
private:
    int res=1;
    map<int,int> mymap; //标记值-位置，并且使用map而不是unordered_map,因为键值有序
};

分析：

我觉得比较好的两点有什么呢：

1.使用map而不是unordered_map，因为键值有序，那么可以结合lower_bound等进行位置查找

2.如何去找一个小于等于当前的位置？考虑先用upper_bound，这样找到了第一个大于当前的，那么当前值又没有进行过记录，因此我们这时候先进行迭代器--，即可保证，所得的就是第一个小于等于的位置。因此，此时需要进行迭代器是否为begin的判断