Come from : [https://leetcode-cn.com/problems/assign-cookies/]
376. Wiggle Subsequence
1.Question
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.
Follow up:
Can you do it in O(n) time?
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
2.Answer
meduim 类型题目。贪心算法(利用状态机比较好理解)。
算法分析:
1.当几个数组出现连续递增时,用贪心算法选择 最大的那个。
2.状态机示意图
其中蓝色部分是 状态机的 保持状态。
红色字 部分 表示 状态机的 转换状态。
3.下图更直观的表示出 各个状态。
AC代码如下:
class Solution {
public:
int wiggleMaxLength(vector<int>& nums) {
if(nums.size() < 2)
{
return nums.size();
}
//扫描序列时的三种状态
static const int BEGIN = 0;
static const int UP = 1;
static const int DOWN = 2;
int STATE = BEGIN;
int max_length = 1; //长度至少为1
for(int i = 1; i < nums.size(); ++i)//从第二个元素开始扫描
{
switch(STATE)
{
case BEGIN:
if(nums[i] > nums[i-1])
{
STATE = UP;
++max_length;
}
else if(nums[i] < nums[i-1])
{
STATE = DOWN;
++max_length;
}
break;
case UP:
if(nums[i] < nums[i-1])
{
STATE = DOWN;
++max_length;
}
break;
case DOWN:
if(nums[i] > nums[i-1])
{
STATE = UP;
++max_length;
}
break;
}
}
return max_length;
}
};
3.大神的算法
大神AC速度第一代码:
很巧妙的做法。
把上面的 状态机 理解了,这个做法应该不难理解。
class Solution {
public:
int wiggleMaxLength(vector<int>& nums) {
int p = 1, q = 1;
for(int i = 1; i < n = nums.size(); ++i){
if(nums[i] > nums[i - 1]) p = q + 1;
if(nums[i] < nums[i - 1]) q = p + 1;
}
return min(n, max(p, q));
}
};
4.我的收获
贪心算法初步。。。
fighting。。。
端午节快乐~记得吃粽子,端午最后一天。。。
2019/6/8 胡云层 于南京 96