975. Odd Even Jump #

题目 #

You are given an integer array arr. From some starting index, you can make a series of jumps. The (1^st, 3^rd, 5^th, …) jumps in the series are called odd-numbered jumps , and the (2^nd, 4^th, 6^th, …) 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 arr[i] <= arr[j] and arr[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 arr[i] >= arr[j] and arr[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 arr.length - 1) by jumping some number of times (possibly 0 or more than once).

Return the number ofgood starting indices.

Example 1:

Input: arr = [10,13,12,14,15]
Output: 2
Explanation:
From starting index i = 0, we can make our 1st jump to i = 2 (since arr[2] is the smallest among arr[1], arr[2], arr[3], arr[4] that is greater or equal to arr[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: arr = [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 arr[1] is the smallest value in [arr[1], arr[2], arr[3], arr[4]] that is greater than or equal to arr[0].
During our 2nd jump (even-numbered), we jump from i = 1 to i = 2 because arr[2] is the largest value in [arr[2], arr[3], arr[4]] that is less than or equal to arr[1]. arr[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 arr[3] is the smallest value in [arr[3], arr[4]] that is greater than or equal to arr[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: arr = [5,1,3,4,2]
Output: 3
Explanation: We can reach the end from starting indices 1, 2, and 4.

Constraints:

• 1 <= arr.length <= 2 * 10^4
• 0 <= arr[i] < 10^5

代码 #

package leetcode

import (
	"fmt"
)

func oddEvenJumps(A []int) int {
	oddJumpMap, evenJumpMap, current, res := map[int]int{}, map[int]int{}, 0, 0
	for i := 0; i < len(A); i++ {
		for j := i + 1; j < len(A); j++ {
			if v, ok := oddJumpMap[i]; ok {
				if A[i] <= A[j] && A[j] <= A[v] {
					if A[j] < A[v] {
						oddJumpMap[i] = j
					}
				}
			} else {
				if A[i] <= A[j] {
					oddJumpMap[i] = j
				}
			}
		}
	}
	for i := 0; i < len(A); i++ {
		for j := i + 1; j < len(A); j++ {
			if v, ok := evenJumpMap[i]; ok {
				if A[i] >= A[j] && A[j] >= A[v] {
					if A[j] > A[v] {
						evenJumpMap[i] = j
					}
				}
			} else {
				if A[i] >= A[j] {
					evenJumpMap[i] = j
				}
			}
		}
	}
	fmt.Printf("oddJumpMap = %v evenJumpMap = %v\n", oddJumpMap, evenJumpMap)
	for i := 0; i < len(A); i++ {
		count := 1
		current = i
		for {
			if count%2 == 1 {
				if v, ok := oddJumpMap[current]; ok {
					if v == len(A)-1 {
						res++
						break
					}
					current = v
					count++
				} else {
					break
				}
			}
			if count%2 == 0 {
				if v, ok := evenJumpMap[current]; ok {
					if v == len(A)-1 {
						res++
						break
					}
					current = v
					count++
				} else {
					break
				}
			}
		}
	}
	return res + 1
}