LeetCode 376. Wiggle Subsequence

376. Wiggle Subsequence

一、问题描述

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.

二、输入输出

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?

三、解题思路

贪心算法

• 养成一个好习惯，首先处理特殊情况：数组长度为0或为1
• 将数组的diff存储到一个单独的数组中，然后遍历diff数组
• 使用贪心算法，尽可能的多的把符合条件的加进来。其实位置需要单独处理，因为diff==0的是不能要的，一直找直到找到第一个不为零的。如果数组结束了，说明所有数都相等，返回1.否则如果相邻两个数是异号的那么count++，并跟新last last中记录的是最终选出来子序列最后一个diff的符号。如果是同号，那么直接跳过该点即可。
• count中是记录的diff中子序列的长度，原序列的长度是count+1
• 判断是不是同号，只需要把两者相乘，如果为正则同号，如果为负，则异号。注意同样不能为零。

class Solution {public:    int wiggleMaxLength(vector<int>& nums) {        if(nums.size() == 0)return 0;        if(nums.size() == 1)return 1;        vector<int> diff;        for (int i = 1, n = nums.size(); i < n; ++i) {            diff.push_back(nums[i] - nums[i-1]);        }        int last, count = 0, first = 0;        while(first < diff.size() && diff[first] == 0) first++;        if(first == diff.size())return 1;        else{            count = 1;            last = diff[first];        }        for (int j = first + 1, n = diff.size(); j < n; ++j) {            if(diff[j] * last < 0){                count++;                last = diff[j];            }        }        return count+1;    }};