LeetCode 376. Wiggle Subsequence
原题链接在这里:https://leetcode.com/problems/wiggle-subsequence/description/
题目:
A sequence of numbers is called a wiggle sequence if 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 fewer than two elements is trivially a wiggle sequence.
For example, [1,7,4,9,2,5]
is a wiggle sequence because the differences (6,-3,5,-7,3) are alternately positive and negative. In contrast, [1,4,7,2,5]
and [1,7,4,5,5]
are not wiggle sequences, the first because its first two differences are positive and the second because its last difference is zero.
Given a sequence of integers, return the length of the longest subsequence that is a wiggle sequence. A subsequence is obtained by deleting some number of elements (eventually, also zero) from the original sequence, leaving the remaining elements in their original order.
Examples:
Input: [1,7,4,9,2,5] Output: 6 The entire sequence is a wiggle sequence. Input: [1,17,5,10,13,15,10,5,16,8] Output: 7 There are several subsequences that achieve this length. One is [1,17,10,13,10,16,8]. Input: [1,2,3,4,5,6,7,8,9] Output: 2
Follow up:
Can you do it in O(n) time?
题解:
up表示到目前为止longest最后是上升的wiggle subsequence长度.
down表示到目前为止longest最后是下降的wiggle subsequence长度.
递推时, 若num[i] > nums[i-1], 就用之前下降的最长wiggle subsequence长度+1.
若nums[i] < nums[i-1], 就用之前上升的最长wiggle subsequence长度+1.
Note: there could be equal number like [0, 0]. The result is 1. Thus it needs to check else if(nums[i] < nums[i - 1]), but not just use else.
Time Complexity: O(nums.length).
Space: O(1).
AC Java:
1 class Solution { 2 public int wiggleMaxLength(int[] nums) { 3 if(nums.length < 2){ 4 return nums.length; 5 } 6 int up = 1; 7 int down = 1; 8 for(int i = 1; i<nums.length; i++){ 9 if(nums[i] > nums[i-1]){ 10 up = down+1; 11 }else if(nums[i] < nums[i-1]){ 12 down = up+1; 13 } 14 } 15 return Math.max(up, down); 16 } 17 }