leetcode376. Wiggle Subsequence

CoffeX 發(fā)布于2019-08-16 17:10 / 3490人閱讀

摘要：題目要求扭動(dòng)序列是指數(shù)組中的相鄰兩個(gè)元素的差保證嚴(yán)格的正負(fù)交替，如數(shù)組中相鄰兩個(gè)元素的差為，滿足扭動(dòng)序列的要求。現(xiàn)在要求從一個(gè)數(shù)組中，找到長(zhǎng)度最長(zhǎng)的扭動(dòng)子序列，并返回其長(zhǎng)度。即前一個(gè)元素和當(dāng)前元素構(gòu)成下降序列，因此代碼如下

題目要求

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?

扭動(dòng)序列是指數(shù)組中的相鄰兩個(gè)元素的差保證嚴(yán)格的正負(fù)交替，如[1,7,4,9,2,5]數(shù)組中相鄰兩個(gè)元素的差為6,-3,5,-7,3，滿足扭動(dòng)序列的要求。現(xiàn)在要求從一個(gè)數(shù)組中，找到長(zhǎng)度最長(zhǎng)的扭動(dòng)子序列，并返回其長(zhǎng)度。

思路和代碼

這是一個(gè)可以通過(guò)動(dòng)態(tài)規(guī)劃來(lái)解決的問(wèn)題。動(dòng)態(tài)規(guī)劃的特點(diǎn)就是，加入我知道第i個(gè)元素的結(jié)果，那么第i+1個(gè)元素的結(jié)果可以由其推到出來(lái)。這里假設(shè)我們知道，以第i個(gè)元素為止的最長(zhǎng)子序列長(zhǎng)度，包括上升序列up和下降序列down，則第i+1個(gè)元素的可能情況如下：

nums[i+1]>nums[i]: 即前一個(gè)元素和當(dāng)前元素構(gòu)成上升序列，因此up[i+1]=down[i]+1, down[i+1]=down[i]，這是指以第i個(gè)元素為結(jié)尾的上升序列應(yīng)當(dāng)基于第i-1個(gè)元素為結(jié)尾的下降序列，而以第i個(gè)元素為結(jié)尾的下降序列，等同于基于第i-1個(gè)元素為結(jié)尾的下降序列。

nums[i+1]>nums[i]: 即前一個(gè)元素和當(dāng)前元素構(gòu)成下降序列，因此down[i+1]=up[i]+1, up[i+1]=up[i]

nums[i+1]=nums[i]: down[i+1]=down[i], up[i+1]=up[i]

代碼如下：

    public int wiggleMaxLength(int[] nums) {
        if( nums.length == 0 ) return 0;
        int[] up = new int[nums.length];
        int[] down = new int[nums.length];
        up[0] = 1;
        down[0] = 1;
        for(int i = 1 ; i nums[i-1]) {
                up[i] = down[i-1] + 1;
                down[i] = down[i-1];
            }else if(nums[i] < nums[i-1]) {
                down[i] = up[i-1] + 1;
                up[i] = up[i-1];
            }else {
                down[i] = down[i-1];
                up[i] = up[i-1];
            }
        }
        return Math.max(up[nums.length-1], down[nums.length-1]);
    }