A wiggle sequence is a sequence where the differences between successive numbers strictly alternate between positive and negative. The first difference (if one exists) may be either positive or negative. A sequence with one element and a sequence with two non-equal elements are trivially wiggle sequences.
- For example,
[1, 7, 4, 9, 2, 5] is a wiggle sequence because the differences(6, -3, 5, -7, 3) alternate between positive and negative. - In contrast,
[1, 4, 7, 2, 5] and[1, 7, 4, 5, 5] are not wiggle sequences. The first is not because its first two differences are positive, and the second is not because its last difference is zero.
A subsequence is obtained by deleting some elements (possibly zero) from the original sequence, leaving the remaining elements in their original order.
Given an integer array
Example 1:
Input: nums = [1,7,4,9,2,5] Output: 6 Explanation: The entire sequence is a wiggle sequence with differences (6, -3, 5, -7, 3).
Example 2:
Input: nums = [1,17,5,10,13,15,10,5,16,8] Output: 7 Explanation: There are several subsequences that achieve this length. One is [1, 17, 10, 13, 10, 16, 8] with differences (16, -7, 3, -3, 6, -8).
Example 3:
Input: nums = [1,2,3,4,5,6,7,8,9] Output: 2
Constraints:
1 <= nums.length <= 1000 0 <= nums[i] <= 1000
Follow up: Could you solve this in
Solution
一開始看到 subsequence,又求極值,直覺會想到 DP
但是推導公式倒不出來,後來發現一個比較簡單的想法
分兩個數列來看,遞增和遞減
因為題目要求要穿插,所以遞增的下個長度會是遞減數列加一
同理,遞減的下個長度會是遞增數列加一
因此,只要每次往前檢查,目前的狀態是遞增還是遞減,對應增加長度就好
kotlin(參考解法)
This file contains bidirectional Unicode text that may be interpreted or compiled differently than what appears below. To review, open the file in an editor that reveals hidden Unicode characters.
Learn more about bidirectional Unicode characters
class Solution { | |
fun wiggleMaxLength(nums: IntArray): Int { | |
var positiveLength = 1 | |
var negativeLength = 1 | |
for(i in 1 until nums.size) { | |
if(nums[i-1] < nums[i]) { | |
positiveLength = negativeLength + 1 | |
} else if(nums[i-1] > nums[i]) { | |
negativeLength = positiveLength + 1 | |
} | |
} | |
return Math.max(positiveLength, negativeLength) | |
} | |
} |
沒有留言:
張貼留言