975. Odd Even Jump

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. Note that the jumps are numbered, not the indices.

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

  • During odd-numbered jumps (i.e., 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 indices j, you can only jump to the smallest such index j.
  • During even-numbered jumps (i.e., 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 indices 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 indices.

Example 1:

Input: A = [10,13,12,14,15]
Output: 2
Explanation:
From starting index i = 0, we can make our 1st 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 cannot jump any more.
From starting index i = 1 and i = 2, we can make our 1st jump to i = 3, then we cannot jump any more.
From starting index i = 3, we can make our 1st jump to i = 4, so we have reached the end.
From starting index i = 4, we have reached the end already.
In total, there are 2 different starting indices i = 3 and i = 4, where we can reach the end with some number of
jumps.

Example 2:

Input: A = [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 indices i = 1, i = 3, and i = 4, where we can reach the end with some
number of jumps.

Example 3:

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

Constraints:

  • 1 <= A.length <= 2 * 104
  • 0 <= A[i] < 105

这道题给了一个数组,可以在任意的位置进行跳跃,分为奇数跳跃和偶数跳跃。第一次跳跃就是奇数跳跃,第二次就是偶数,第三次又是奇数,以此类推。奇数跳跃时到达的位置上的数字必须要大于等于起跳位置的数字,若有多个位置的数字都大于等于起跳位置,选其中最小的,若数字相同,选坐标最小的。而偶数跳跃到达的位置上的数字必须要小于等于起跳位置的数字,若有多个位置的数字都小于等于起跳位置,选其中最大的,若数字相同,选坐标最小的。现在定义了一种好起点,需要能按照上面的跳跃方式到达数组最后一个位置,问有多少个这样的好起点。说实话,这道题的题目博主是看了好久才搞懂,看到论坛上也有人吐槽题意晦涩难懂的。不过点赞数远超踩的个数,看来还是一道不错的题目。由于起点是任意的,那么若起点就是在最后一个位置,则就不用跳了,所以结果 res 可以初始化为1。然后就可以往前推,对于前一个数字和当前数字的关系,实际上就是大于等于,或者小于等于,可以分别对应两种跳法,这样其实每个位置上就有两种状态,一种是能否跳到大于等于的位置,用 higher 表示,一种是能否跳到小于等于的位置,用 lower 表示。这样就可以用两个数组 higher 和 lower 表示,其中 higher[i] 表示起点为i位置,首先跳到大于等于的位置(奇数跳跃),看是否能跳到末尾位置,这个就是题目所要求的。lower[i] 表示起点为i位置,首先跳到小于等于的位置(偶数跳跃),看是否能跳到末尾位置。则最末尾的位置 higher[n-1] 和 lower[n-1] 都要初始化为 true。在往前推的时候,需要在后方的数字中找出第一个不小于当前数字的数,和第一个不大于当前数字的数,为了快速查找,可以使用 TreeMap 来建立数字和其下标之间的映射,然后就可以用 lower_bound 和 upper_bound 来快速的查找了。这里 lower_bound 是查找第一个不小于目标值的数,正好就是要求的,只要该数字存在,则可以用该数字的 lower 值来更新当前数字的 higher 值,因为奇数跳跃和偶数跳跃是要交替进行的。这里的 upper_bound 是查找第一个大于目标的数,其往前退一位就是第一个不大于目标的数,但是在退之前,要先确定这不是第一个数字,否则没法往前退。用查找到的数字的 higher 值来更新当前数字的 lower 值。每次若 higher 值为 true,则结果 res 自增1,参见代码如下:

class Solution {
public:
    int oddEvenJumps(vector<int>& A) {
        int res = 1, n = A.size();
        vector<bool> higher(n), lower(n);
        higher[n - 1] = lower[n - 1] = true;
        map<int, int> num2idx;
        num2idx[A[n - 1]] = n - 1;
        for (int i = n - 2; i >= 0; --i) {
            auto hi = num2idx.lower_bound(A[i]), lo = num2idx.upper_bound(A[i]);
            if (hi != num2idx.end()) higher[i] = lower[hi->second];
            if (lo != num2idx.begin()) lower[i] = higher[(--lo)->second];
            if (higher[i]) ++res;
            num2idx[A[i]] = i;
        }
        return res;
    }
};

Github 同步地址:

https://github.com/grandyang/leetcode/issues/975

参考资料:

https://leetcode.com/problems/odd-even-jump/

https://leetcode.com/problems/odd-even-jump/discuss/217974/Java-solution-DP-%2B-TreeMap

https://leetcode.com/problems/odd-even-jump/discuss/217981/JavaC%2B%2BPython-DP-using-Map-or-Stack

LeetCode All in One 题目讲解汇总(持续更新中…)


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