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.

Example 1:

Input: [1,7,4,9,2,5]
Output: 6
Explanation: The entire sequence is a wiggle sequence.

Example 2:

Input: [1,17,5,10,13,15,10,5,16,8]
Output: 7
Explanation: There are several subsequences that achieve this length. One is [1,17,10,13,10,16,8].

Example 3:

Input: [1,2,3,4,5,6,7,8,9]
Output: 2

Follow up:

Can you do it in O(n) time?

方法1: two pointers （错了！一定要分清subsequence 和 subarray的区别）

思路：

遍历第一遍得出一个sign的vector。第二遍右指针开始向右遍历检查是否有nums[i - 1] 和nums[i]相同，如果有：1. 更新result，2. 左指针接替右指针，3. 右指针后移。

class Solution {
public:
    int wiggleMaxLength(vector<int>& nums) {
        if (nums.size() <= 1){ return nums.size();}
        if (nums.size() == 2) {return nums[0] == nums[1] ? 1 : 2;}
        int left = 1, right = 2;
        int prevSign = nums[1] - nums[0] > 0 ? 1 : -1;
        int curSign;
        int result = 0;
        while (right < nums.size()){
            int diff = nums[right] - nums[right - 1];
            if (prevSign * diff >= 0 ){
                left = right;
            }
            else {
                result = max(result, right - left + 2);
            }
            prevSign = diff > 0 ? 1 : -1;
            right ++;
            
        }
        return result;
    }
};

方法1: dynamic programming

思路：

用两个dp记录前面到第i项结束的wiggle subseq 长度，和当前比较。如果diff > 0, dpPos[i] = dpNeg[i - 1] + 1，dpNeg[i] = dpNeg[i - 1]. 如果diff < 0, 那么dpNeg[i] = dpPos[i - 1] + 1, dpPos[i] = dpPos[i - 1]. if diff == 0, dpPos[i] = dpPos[i - 1], dpNeg[i] = dpNeg[i - 1]

Complexity

Time complexity: O(n)
Space complexity: O(n)

易错点

1. (wiggle dp)
2. initialization = 1，for convenience

class Solution {
public:
    int wiggleMaxLength(vector<int>& nums) {
        if (nums.size() <= 1){ return nums.size();}
        if (nums.size() == 2) {return nums[0] == nums[1] ? 1 : 2;}
        vector<int> dpPos(nums.size() , 1);
        vector<int> dpNeg(nums.size() , 1);
        
        for (int i = 1; i < nums.size(); i ++){
            int diff = nums[i] - nums[i - 1];
            if (diff > 0){
                dpNeg[i] = dpPos[i - 1] + 1;
                dpPos[i] = dpPos[i - 1];
            }
            else if (diff < 0){
                dpPos[i] = dpNeg[i - 1] + 1;
                dpNeg[i] = dpNeg[i - 1];
            }
            else {
                dpPos[i] = dpPos[i - 1];
                dpNeg[i] = dpNeg[i - 1];
            }
        }
        return max(dpPos.back(), dpNeg.back());
    }
};
//          [1,17,5,10,13,15,10,5,16,8]
//   
// dpPos     1 2  2 4  4  4  4  4  6 6 
// dpNeg     1 1  3 3  3  3  5  5  5 7

//          [1, 7, 4, 5, 5]
// dpPos     1  2  2  4  4
// dpNeg     1  1  3  3  3

//          [1, 7, 4, 9, 2, 5]
// dpPos     1  2  2  4  4  6
// dpNeg     1  1  3  3  5  5

方法2: dynamic programming，one-dimension/greedy

思路：

降维之后，省略一些步骤，不知道为什么就成greedy了。另外这个方法很巧妙的利用min(n, max(dpPos, dpNeg)) 来解决了上面0， 1，2 长度的所有edge cases。。。。！

Complexity

Time complexity: O(n)
Space complexity: O(1)

class Solution {
public:
    int wiggleMaxLength(vector<int>& nums) {
        int n = nums.size();
        int dpPos = 1;
        int dpNeg = 1;
        
        for (int i = 1; i < nums.size(); i ++){
            int diff = nums[i] - nums[i - 1];
            if (diff > 0){
                dpNeg = dpPos + 1;
            }
            else if (diff < 0){
                dpPos = dpNeg + 1;
            }
        }
        return min(n, max(dpPos, dpNeg));
    }
};