Leetcode学习日志-376 Wiggle Subsequence

Leetcode 376 Wiggle Subsequence

题目原文

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?

题意分析

从一串整数中找出蛇形排列的字串，求出最长的蛇形字串，对于长度为0和1的串，认为蛇形字串长度为1。本题采用暴力方法时间复杂度为O(n!)，所以需要采用巧妙的解法。

解法分析

本题采用两种解法：

• 动态规划
• 贪心算法

动态规划

本题采用自底向上的动态规划方法，分别创建数组up[n]，down[n]来存储遍历到字串的第i个元素时，最后一次是up和down的蛇形字串长度，示例如下：

C++代码如下：

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

上述算法运算时间复杂度为O(n)，为了缩减空间复杂度，由于每次计算只需用到前一次的up和down，所以不需要用数组存储所有值，只需用一个变量存储。

贪心算法

本题不需要用DP，通过分析可以看到，当数一直增加或一直减少时，总是需要选取最大或最小的那个元素进入蛇形串，示意图如下：

可以看到A、B、C、D都在上升，选A之后应该选D，这样可以保证后来得到的蛇形串最长，因为它有更大概率遇到比它小的数，形成蛇形串，这也是贪心算法正确的原因。C++代码如下：

class Solution {public:    int wiggleMaxLength(vector<int>& nums) {        int n=nums.size();        int i;        if(n==0||n==1)            return n;        int sum=1;        int test=2;//Use test to check the up and down        for(i=1;i<n;i++){            if(test==2){                if(nums[i]>nums[i-1])                    test=1;                else if(nums[i]<nums[i-1])                    test=0;                 if(test==2)//the input is[0,0]or[1,1]                    continue;                sum++;                continue;            }            if((test==1&&(nums[i]<nums[i-1]))||(test==0&&(nums[i]>nums[i-1]))){                test=1-test;                sum++;                continue;              }           }        return sum;       }};

上述方法更简单，关键是寻找到蛇形串的特点。