[解题报告]376. Wiggle Subsequence

Problem Background:

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?

Credits:
Special thanks to @agave and @StefanPochmann for adding this problem and creating all test cases.

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 Dynamic Programming Greedy

解体报告：

从Tag可以看出这是一道DP＋贪心解法

O(n)时间

1.线性扫描整个数据集

2.维护一个lastNum值，它纪录了上一次连续非单调的最值(最小值或者最大值)

3.每次迭代，和lastNum比较，如果满足wiggle，max加一

Code：

class Solution {public:    int wiggleMaxLength(vector<int>& nums) {        int length = nums.size();        if (length < 1) {        return 0;        } else if (length == 1) {        return 1;        } else {        int max = 1;        int lastNum = nums[0];        bool isPositive = nums[1] > nums[0] ? true : false;        for (int i = 1; i < length; ++i, isPositive = !isPositive) {        if (isPositive) {        if (nums[i] > lastNum) {        ++max;        } else {        isPositive = false;        }        } else {        if (nums[i] < lastNum) {        ++max;        } else {        isPositive = true;        }        }lastNum = nums[i];        }        return max;    }    }};