376 Wiggle Subsequence
376. Wiggle Subsequence
1. Question
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?
2. Implementation
(1) DP
class Solution {
public int wiggleMaxLength(int[] nums) {
if (nums.length <= 1) {
return nums.length;
}
int n = nums.length;
int[][] dp = new int[2][n];
Arrays.fill(dp[0], 1);
Arrays.fill(dp[1], 1);
int maxLen = 1;
for (int i = 1; i < n; i++) {
for (int j = 0; j < i; j++) {
if (nums[j] < nums[i] && dp[1][j] + 1 > dp[0][i]) {
dp[0][i] = dp[1][j] + 1;
}
else if (nums[j] > nums[i] && dp[0][j] + 1 > dp[1][i]) {
dp[1][i] = dp[0][j] + 1;
}
}
maxLen = Math.max(maxLen, Math.max(dp[0][i], dp[1][i]));
}
return maxLen;
}
}
3. Time & Space Complexity
(1) DP: 时间复杂度O(n^2), 空间复杂度O(n)
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