Leetcode-376Wiggle Subsequence

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?

题解：

一个整数序列，如果两个相邻元素的差恰好正负（负正）交替出现，则该序列被称为摇摆序列；小于两个元素的序列直接为摇摆序列；
本题输入一个序列，求出该序列中，满足摇摆序列定义的最长子序列的长度。
我们以 [1,2,7,4,3,9,2,2,5]为例：
满足摇摆序列定义的最长子序列为：[1,7,3,9,2,5]
贪心的思想来解决这个问题：
从元素1开始，相邻的两元素依次进行比较；我们发现，元素2比1大，元素7比2大，摇摆序列要求满足正负（负正）交替出现，明显取最大的元素7时更容易得到更多的比7小的数；所以当元素递增时，取最大的元素是最优的策略；
同理，当元素递减时，取最小的元素是最优的策略，因为这样可以尽可能多的获得比该元素大的数；
思路确定了，我们要怎么统计最长子序列的长度呢？我们发现，因为所得到的摇摆序列肯定是满足正负（负正）交替出现的，所以当元素递增状态切换到元素递减状态时，增加了一个子序列元素；所以当元素递减状态切换到元素递增状态时，增加了一个子序列元素；
基于以上规律，我们决定使用状态机来解决统计子序列长度的问题；

image.png
如图，对于序列[1,2,7,4,3,9,2,2,5]：
count = 0：start(初始状态)：元素1
count = 1：start(初始状态)->up(递增状态)：元素1到元素2
count = 2：up(递增状态)->down(递减状态)：元素7到元素4
count = 3：down(递减状态)->up(递增状态)：元素3到元素9
count = 4：up(递增状态)->down(递减状态)：元素9到元素2
count = 5：down(递减状态)->up(递增状态)：元素2到元素5
状态总共切换了5次，总共有6个元素，序列为[1,7,3,9,2,5]；
注：相邻元素相等时不会影响状态变化，无视即可；

My Solution（C/C++完整实现）：

#include <cstdio>
#include <iostream>
#include <vector>

using namespace std;

class Solution {
public:
    int wiggleMaxLength(vector<int> &nums) {
        int count = 1;
        int state = 0;
        //int value = nums[0];
        for (int i = 1; i < nums.size(); i++) {
            if (nums[i - 1] < nums[i]) {
                if (state != 1) {
                    count += 1;
                    state = 1;
                }
            }
            else if (nums[i - 1] > nums[i]) {
                if (state != -1) {
                    count += 1;
                    state = -1;
                }
            }
        }
        return count;
    }
};

int main() {
    Solution s;
    vector<int> nums;
    nums.push_back(1);
    nums.push_back(7);
    nums.push_back(4);
    nums.push_back(9);
    nums.push_back(9);
    nums.push_back(2);
    nums.push_back(5);
    printf("%d\n", s.wiggleMaxLength(nums));
    getchar();
    return 0;
}

结果：

6

My Solution（Python）：

class Solution:
    def wiggleMaxLength(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        if len(nums) == 0:
            return 0
        state = 0
        judge_num = nums[0]
        result = 1
        # min_num = nums[0]
        for i in range(1, len(nums)):
            if nums[i] > judge_num:
                judge_num = nums[i]
                if state != 1:
                    result += 1
                    state = 1
            if nums[i] < judge_num:
                judge_num = nums[i]
                if state != -1:
                    result += 1
                    state = -1
        return result

Reference:

class Solution:
    def wiggleMaxLength(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        p, q = 1, 1
        for i in range(1, len(nums)):
            if nums[i] > nums[i-1]:
                p = q + 1
            if nums[i] < nums[i-1]:
                q = p + 1
        return min(max(p, q), len(nums))