376. Wiggle Subsequence

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


Code

public class Solution {
    public class Data {
        public int pre;
        public int count;
        
        public Data(int index) {
            this.pre = index;
            this.count = 1;
        }
    }
    
    public int wiggleMaxLength(int[] nums) {
        if (nums.length <= 1) return nums.length;
        Data[] dp = new Data[nums.length];
        dp[0] = new Data(0);
        
        int max = 1;
        for (int i = 1; i < dp.length; i++) {
            dp[i] = new Data(i);
            for (int j = 0; j < i; j++) {
                if (nums[j] != nums[i]) {
                    if (dp[j].pre == j) {
                        dp[i].count = Math.max(dp[i].count, 1 + dp[j].count);
                        dp[i].pre = j;
                        continue;
                    }
                    
                    int first = nums[dp[j].pre], second = nums[j], third = nums[i];
                    if ((second - first < 0 && third - second > 0) || (second - first > 0 && third - second < 0)) {
                        dp[i].count = Math.max(dp[i].count, 1 + dp[j].count);
                        dp[i].pre = j;
                    }
                }
            }
            max = Math.max(max, dp[i].count);
        }
        
        return max;
    }
}

Solution

动态规划 + 贪心。

自定义一个内部类,包含2个元素:pre指序列中当前元素的前一个元素在nums中的下标,count指以当前元素结尾的序列的长度。

dp[i].count表示以nums[i]结尾的序列的最长长度。
初始化时,dp[0].pre = 0, count = 1;

状态转移方程:dp[i].count = Math.max(dp[i].count, 1 + dp[j].count);
注意可以转移的条件即可。

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