376. Wiggle Subsequence
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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: 6The entire sequence is a wiggle sequence.Input: [1,17,5,10,13,15,10,5,16,8]Output: 7There 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?
典型的贪心算法,找一个小或者大的找到最后一个,因为如果不是互换的就是一直递增或者递减的,那么最后一个就必然是最大或者最小的,那么要删除也是删除前面的,就可以得到最长的交叉数列。但最开始可能是从大于或者是从小于开始,所以要两次判定,找较大值。复杂度是O(n)。
实现代码如下:
class Solution {public: int wiggleMaxLength(vector<int>& nums) { if(nums.size()==0) return 0; int count = 1,count2 = 1; bool alt = 0,alt1 = 1; for(int i=1;i<nums.size();i++) { if(((nums[i]>nums[i-1])^alt)&&nums[i]!=nums[i-1]) { count++; alt=!alt; } if(((nums[i]>nums[i-1])^alt1)&&nums[i]!=nums[i-1]) { count2++; alt1=!alt1; } } return count>count2? count: count2; }};
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