[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.

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?

解法一：

dp的思想。up，down两个数组的元素分别表示到第i个数字，以上升或者下降结束的最长subsequence的长度。nums[i] > nums[i-1]表示上升，down[i-1]是在i-1元素以下降结束的subsequence的长度，所以up[i] = down[i-1] + 1, down[i] = down[i-1]。相应的得到nums[i] < nums[i-1]的递推公式。 

class Solution {
public:
    int wiggleMaxLength(vector<int>& nums) {
        if(nums.size()<=2) return nums.size(); 
        vector<int> up(nums.size(),0);
        vector<int> down(nums.size(),0);
        up[0] = down[0] = 1;
        
        for(int i=1; i<nums.size(); i++){
            if(nums[i-1]<nums[i]){
                up[i] = down[i-1]+1;
                down[i] = down[i-1];
            }
            else if(nums[i-1]>nums[i]){
                down[i] = up[i-1]+1;
                up[i] = up[i-1];
            }
            else{
                down[i] = down[i-1];
                up[i] = up[i-1];
            }
        }
        return max(down.back(), up.back());
        
        
    }
};

解法二：

O(n)的方法。up表示以nums[i-1]结尾并且上升的最长subsequence的长度，down表示以nuts[i-1]结尾并且下降的最长subsequence的长度。所以，当num[i]>nums[i-1]说明可以接上down的subsequence，up长度为down+1。如果num[i] < nums[i-1]，则down = up + 1。

class Solution {
public:
    int wiggleMaxLength(vector<int>& nums) {
        int up = 1, down = 1, n = nums.size();
        for(int i= 1; i<n; i++){
            if (nums[i]>nums[i-1]) up = down + 1;
            else if (nums[i] < nums[i-1]) down = up + 1;
        }
        return min(n,max(up,down));
        
    }
};