[LeetCode] 376. Wiggle Subsequence

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.

Example 1:

Input: [1,7,4,9,2,5]
Output: 6
Explanation: The entire sequence is a wiggle sequence.

Example 2:

Input: [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].

Example 3:

Input: [1,2,3,4,5,6,7,8,9]
Output: 2

Follow up:
Can you do it in O(n) time?

摆动序列。

如果连续数字之间的差严格地在正数和负数之间交替，则数字序列称为摆动序列。第一个差（如果存在的话）可能是正数或负数。少于两个元素的序列也是摆动序列。

例如， [1,7,4,9,2,5] 是一个摆动序列，因为差值 (6,-3,5,-7,3) 是正负交替出现的。相反, [1,4,7,2,5] 和 [1,7,4,5,5] 不是摆动序列，第一个序列是因为它的前两个差值都是正数，第二个序列是因为它的最后一个差值为零。

给定一个整数序列，返回作为摆动序列的最长子序列的长度。 通过从原始序列中删除一些（也可以不删除）元素来获得子序列，剩下的元素保持其原始顺序。

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/wiggle-subsequence
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

思路是动态规划。这道题的思路很像300题最长上升子序列，我们创建两个数组up和down，表示以nums[i]为结尾的，最后是上升/下降的最长子序列的长度。

如果nums[i] > nums[i - 1]，说明最后是上升的，最长子序列的长度是down[i - 1] + 1，因为最后是上升的话，再往前应该是下降的

如果nums[i] < nums[i - 1]，说明最后是下降的，最长子序列的长度是up[i - 1] + 1，因为最后是下降的话，再往前应该是上升的

如果nums[i] == nums[i - 1]，说明已经破坏了上升/下降的规则，则最长子序列的长度跟前一个位置相同

时间O(n)

空间O(n)

Java实现

class Solution {
    public int wiggleMaxLength(int[] nums) {
        // corner case
        if (nums.length <= 1) {
            return nums.length;
        }

        // normal case
        int len = nums.length;
        int[] down = new int[len];
        int[] up = new int[len];
        down[0] = 1;
        up[0] = 1;
        for (int i = 1; i < len; i++) {
            if (nums[i] > nums[i - 1]) {
                down[i] = down[i - 1];
                up[i] = down[i - 1] + 1;
            } else if (nums[i] < nums[i - 1]) {
                down[i] = up[i - 1] + 1;
                up[i] = up[i - 1];
            } else {
                down[i] = down[i - 1];
                up[i] = up[i - 1];
            }
        }
        return Math.max(down[len - 1], up[len - 1]);
    }
}

 

另外提供一个不使用额外空间的做法。

class Solution {
    public int wiggleMaxLength(int[] nums) {
        // corner case
        if (nums.length <= 1) {
            return nums.length;
        }

        // normal case
        int up = 1;
        int down = 1;
        for (int i = 1; i < nums.length; i++) {
            if (nums[i] > nums[i - 1]) {
                up = down + 1;
            } else if (nums[i] < nums[i - 1]) {
                down = up + 1;
            }
        }
        return Math.max(up, down);
    }
}

 

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