376. Wiggle Subsequence

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最新推荐文章于 2018-11-16 17:30:00 发布

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原文链接：http://www.cnblogs.com/neweracoding/p/5707087.html

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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?

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Dynamic Programming Greedy

Solution: the last element in a continuous up/down series will be used as a turn point.

public class Solution {
    public int wiggleMaxLength(int[] nums) {
        if(nums == null)
            return 0;
        if(nums.length <=1)
            return nums.length;
            
        int second = 1; //find the first number that is not equal to nums[0] 
        while(second < nums.length) {
            if(nums[second] != nums[0])
                break;
            ++second;
        }
        if(second == nums.length)
            return 1; //every number is the same
    
        int prevDiff = nums[second] - nums[0]; 
        int previous = second;
        int len = 2;
        for(int i = second+1; i<nums.length; ++i) {
            int diff = nums[i] - nums[previous];
            if(diff == 0)
                continue;
            if((diff<0&&prevDiff>0) || (diff>0&&prevDiff<0)) { //if there is a turn.
                ++len;
                prevDiff = diff;
            }
            previous = i;
        }
        return len;
    }
}