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: 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?
class Solution {public: int wiggleMaxLength(vector<int>& nums) { int n = nums.size(); if(n == 0) return 0; if(n == 1) return 1; int ans = -1; while(!dq1.empty()) dq1.pop_back(); while(!dq2.empty()) dq2.pop_back(); memset(lmax, -1, sizeof(lmax)); memset(lmin, -1, sizeof(lmin)); memset(clmax, -1, sizeof(clmax)); memset(clmin, -1, sizeof(clmin)); memset(dp, -1, sizeof(dp)); dq1.push_back(0); dq2.push_back(0); for(int i = 1; i < n; ++i) { while(!dq1.empty() && nums[dq1.back()] >= nums[i]) dq1.pop_back(); if(!dq1.empty()) lmin[i] = dq1.back(); dq1.push_back(i); while(!dq2.empty() && nums[dq2.back()] <= nums[i]) dq2.pop_back(); if(!dq2.empty()) lmax[i] = dq2.back(); dq2.push_back(i); } clmin[0] = clmax[0] = 0; for(int i = 1; i < n; ++i) { if(nums[i] >= nums[i-1]) clmin[i] = clmin[i-1]; else clmin[i] = i; if(nums[i] <= nums[i-1]) clmax[i] = clmax[i-1]; else clmax[i] = i; } for(int i = 0; i < n; ++i) { if(lmin[i] == -1) { dp[i][1] = 1; } else { dp[i][1] = dp[clmin[lmin[i]]][0] + 1; } if(lmax[i] == -1) { dp[i][0] = 1; } else { dp[i][0] = dp[clmax[lmax[i]]][1] + 1; } ans = max(ans, dp[i][0]); ans = max(ans, dp[i][1]); } return ans; }private: deque<int> dq1; deque<int> dq2; int lmax[100010]; int lmin[100010]; int clmin[100010]; int clmax[100010]; int dp[100010][2];};
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