leetcode 659连续数组分割子连续序列

You are given an integer array sorted in ascending order (may contain duplicates), you need to split them into several subsequences, where each subsequences consist of at least 3 consecutive integers. Return whether you can make such a split.

Example 1:
Input: [1,2,3,3,4,5]
Output: True
Explanation:
You can split them into two consecutive subsequences :
1, 2, 3
3, 4, 5
Example 2:
Input: [1,2,3,3,4,4,5,5]
Output: True
Explanation:
You can split them into two consecutive subsequences :
1, 2, 3, 4, 5
3, 4, 5
Example 3:
Input: [1,2,3,4,4,5]
Output: False
Note:
The length of the input is in range of [1, 10000]
题解：用一个字典left,left[i]表示元素i还剩下多少，end[i]表示以元素i结尾的子序列有多少个，当这个元素既没办法加到前面一个元素的后面，也没法在后面找到两个值构成一个序列的时候就返回false
python code:
def isPossible(self, nums):
“”"
:type nums: List[int]
:rtype: bool
“”"
left,end=collections.Counter(nums),collections.Counter()
for i in nums:
if not left[i]:continue
left[i]-=1
if end[i-1]>0:
end[i-1],end[i]=end[i-1]-1,end[i]+1
elif left[i+1] and left[i+2]:
end[i+2],left[i+1],left[i+2]=end[i+2]+1,left[i+1]-1,left[i+2]-1
else:
return False
return True

**c++Code:**
    bool isPossible(vector& nums) {
    map left,end;
    for(vector::iterator it=nums.begin();it!=nums.end();it++) ++left[*it];
    for(vector::iterator i=nums.begin();i!=nums.end();i++)
    {
        if(!left[*i])continue;
        --left[*i];
        if(end[*i-1]>0)
        {
            --end[*i-1];
            ++end[*i];
        }
        else if(left[*i+1]&&left[*i+2])
        {
            --left[*i+1];
            --left[*i+2];
            ++end[*i+2];
        }
        else
            return false;
    }
    return true;
}