Leetcode 376. Wiggle Subsequence

问题描述

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.

保存两个数组，一行表示当前数作为最后一个波谷时序列长度，一行表示当前数作为最后一个波峰时的序列长度，所以有，如果当前数比之前的数大，则当前数波峰序列数目为之前数波谷序列数目加1，如果当前数比之前的数小，则当前数波谷序列数目为之前数波峰序列数目加1。
O(n2)代码：

    public int wiggleMaxLength(int[] nums) {        int n=nums.length;        if(n==0)            return 0;        int[][]dp=new int[2][n];        dp[0][0]=1;        dp[1][0]=1;        int result=1;        for(int i=1;i<n;i++){            dp[0][i]=1;            dp[1][i]=1;            for(int j=0;j<i;j++){                if(nums[i]>nums[j]){                    dp[0][i]=Math.max(dp[0][i],dp[1][j]+1);                }                if(nums[i]<nums[j]){                    dp[1][i]=Math.max(dp[1][i],dp[0][j]+1);                }            }            if(result<dp[0][i])                result=dp[0][i];            if(result<dp[1][i])                result=dp[1][i];        }        return result;    }

由于每次更新当前数的波峰波谷时必须向前遍历，找到最大的那个可以更新的前一个数，导致算法的复杂度上升达到O(n2)。如果可以将“最大的那个可以更新的前一个数的波峰波谷”向后实时传递，则可以降低复杂度。此时，dp[0][i]表示前（i+）个数最长波动序列的最后一个数是波峰时其序列长度，同理dp[1][i]表示波谷序列长度。

    public int wiggleMaxLength(int[] nums) {        int n=nums.length;        if(n==0)            return 0;        int[][]dp=new int[2][n];        dp[0][0]=1;        dp[1][0]=1;        for(int i=1;i<n;i++){            if(nums[i]>nums[i-1]){                dp[0][i]=dp[1][i-1]+1;                dp[1][i]=dp[1][i-1];            }            else if(nums[i]<nums[i-1]){                dp[0][i]=dp[0][i-1];                dp[1][i]=dp[0][i-1]+1;            }            else{                dp[0][i]=dp[0][i-1];                dp[1][i]=dp[1][i-1];            }        }        return dp[0][n-1]>dp[1][n-1]?dp[0][n-1]:dp[1][n-1];    }

dp[i]只与dp[i-1]有关，所以可以只用2个变量来降低空间复杂度。代码忽略
参考链接：http://bookshadow.com/weblog/2016/07/21/leetcode-wiggle-subsequence/