Description:
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:
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.
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.
Input: [5,1,3,4,2]
Output: 3
Explanation:
We can reach the end from starting indexes 1, 2, and 4.
Link:
https://leetcode.com/problems/odd-even-jump/
解题方法：
DP:
- 由两个数组odd和even来代表在i这个位置上以odd jump和even jump，能不能跳到终点。
- 从后往前递推，最后一个点(终点)odd和even都是True.
- 在i点这个位置，如果能实现odd jump，那么
odd[i] = even[此次odd jump的下一个点位置]；如果能实现even jump，那么even[i] = odd[此次even jump的下一个点位置]. 这样结果就可以一步步传给前面的点。 - 经过以上三步，唯一的问题就是怎么在最短的时间内找到odd 和 even jump的下一个点了，如果采用遍历的方法，那么一次找寻就要花费n的时间。如果使用tree map, 一次找寻只需要花费logn的时间：
- C++中，tree map对应的数据结构是map.
-
low_bound(key)这个函数会返回>= key的迭代器，这就可以得到odd jump的查找结果了。 -
upper_boud(key)这个函数会返回> key的迭代器，那么把这个迭代器--就可以得到even jump的查找结果了。
Time Complexity：
O(nlogn)
完整代码：
int oddEvenJumps(vector& A) {
int size = A.size();
int ans = 1;
vector odd(size, false);
vector even(size, false);
odd[size - 1] = even[size - 1] = true;
map m;
m[A[size - 1]] = size - 1;
for(int i = size - 2; i >= 0; i--) {
auto up_bound = m.lower_bound(A[i]);
auto low_bound = m.upper_bound(A[i]);
if(up_bound != m.end())
odd[i] = even[up_bound->second];
if(low_bound != m.begin())
even[i] = odd[(--low_bound)->second];
//upper_bound is ">" this key, !upper_bound = --upper_bound = "<=" this key
if(odd[i])
++ans;
m[A[i]] = i;
}
return ans;
}