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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