Leetcode 376. Wiggle Subsequence

376. Wiggle Subsequence

• Total Accepted: 5157
• Total Submissions: 14616
• Difficulty: Medium

 

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?

 

思路：DP和贪心。假设返回值是res。如果nums[0]<nums[1]（递增），并且nums[2]<nums[1]（递减），那么res+=1。但如果nums[2]>=nums[1]，那么保留pre=nums[2]（这里可以这么想：之后要的是递减的状态，nums[2]>=nums[1]，可以认为nums[2]比nums[1]接下来更有形成递减状态的可能，所以保留pre=nums[2]）。后面的数只要和pre（nms[2]）比较就可以了。

如果nums[0]>nums[1]的情况也是相似的。

 

代码：

class Solution {
public:
    int wiggleMaxLength(vector<int>& nums) {
        int n=nums.size();
        if(n<2) return n;
        int i,pre=nums[0],res=1;
        bool increasing=nums[0]<nums[1];
        for(i=1;i<n;i++){
            if(increasing&&pre<nums[i]){
                increasing=!increasing;
                res++;
            }
            else{
                if(!increasing&&pre>nums[i]){
                    increasing=!increasing;
                    res++;
                }
            }
            pre=nums[i];
        }
        return res;
    }
};