leetcode659. Split Array into Consecutive Subsequences
leetcode659. Split Array into Consecutive Subsequences
题意:
您将获得按升序排列的整数数组(可能包含重复项),您需要将它们拆分成多个子序列,其中每个子序列至少包含3个连续的整数。返回是否可以进行这样的拆分。
思路:
虽然是medium,但是我感觉有点难想= =。 O(n)复杂度。
dp。
先分开各种不同的序列。按如果间隔大于1就坑定是不同的序列。按此分开不同的序列以此简化问题。
然后处理连续的序列。连续的序列false的情况就是没有足够的数字构成连续的序列。
然后转化成一个mp,值为这个数字的个数。
遍历时用一个one记录以此处为end的序列长度为1的序列。用two记录到此处长度为2的序列。因为贪心的缘故。每次的移动后。因为要尽可能完成一个序列,如果此时这个数字的个数 < one + two的个数。说明这个数字是不够完成one,two要构成连续序列的个数的。所以返回False。two就是上一个的one的值。然后one的值,如果比上个数字的个数大,这个数字的个数和上个数字个数的差。但是如果比上个数字小,那就取零。为了方便,用tot来储存上一个数字的个数。tot表示的含义其实就是要构成序列。下一个数字所必须的长度。
ac代码:
C++
class Solution {
public:
bool isPossible(vector<int>& nums) {
int len = nums.size();
int k = 0;
for(int i = 0; i < len; i++)
{
if(i > 0 && nums[i] > nums[i - 1] + 1)
{
if(!check(nums,k,i - 1)) return false;
else k = i;
}
}
return check(nums,k,len - 1);
}
private:
bool check(vector<int>& nums,int s, int e)
{
int n = nums[e] - nums[s] + 1;
vector<int> mp(n,0);
for(int i = s; i <= e; i++)
{
mp[nums[i] - nums[s]]++;
}
int one, two, tot;
one = two = tot = 0;
for(int i = 0; i < n; i++)
{
if(mp[i] < one + two) return false;
two = one;
one = max(0, mp[i] - tot);
tot = mp[i];
}
return one == 0 && two == 0;
}
};
python
class Solution:
def isPossible(self, nums):
"""
:type nums: List[int]
:rtype: bool
"""
last = 0
for i , num in enumerate(nums):
if i > 0 and nums[i] > nums[i - 1] + 1:
if self.check(nums, last, i - 1) == False:
return False
else:
last = i
return self.check(nums, last, len(nums) - 1)
def check(self, nums, s, e):
mp = []
for i in range(nums[e] - nums[s] + 1):
mp.append(0)
for i in range(s,e + 1):
mp[nums[i] - nums[s]] += 1
one = two = tot = 0
for m in mp:
if m < one + two:
return False
two = one
one = max(0, m - tot)
tot = m
return one == 0 and two == 0
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