[LeetCode] 659. Split Array into Consecutive Subsequences 将数组分割成连续子序列

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:

1. The length of the input is in range of [1, 10000]

给一个升序排列的整数数(可能含有重复元素)，把这个数组拆分成几个至少含有3个整数的子序列，求是否可以拆分成功？（也就是能够全部拆分组成顺子）

解法：贪婪算法Greedy

Java:

public boolean isPossible(int[] nums) {
    Map freq = new HashMap<>(), appendfreq = new HashMap<>();
    for (int i : nums) freq.put(i, freq.getOrDefault(i,0) + 1);
    for (int i : nums) {
        if (freq.get(i) == 0) continue;
        else if (appendfreq.getOrDefault(i,0) > 0) {
            appendfreq.put(i, appendfreq.get(i) - 1);
            appendfreq.put(i+1, appendfreq.getOrDefault(i+1,0) + 1);
        }   
        else if (freq.getOrDefault(i+1,0) > 0 && freq.getOrDefault(i+2,0) > 0) {
            freq.put(i+1, freq.get(i+1) - 1);
            freq.put(i+2, freq.get(i+2) - 1);
            appendfreq.put(i+3, appendfreq.getOrDefault(i+3,0) + 1);
        }
        else return false;
        freq.put(i, freq.get(i) - 1);
    }
    return true;
}

Python:

# Time:  O(n)
# Space: O(1)
class Solution(object):
    def isPossible(self, nums):
        """
        :type nums: List[int]
        :rtype: bool
        """
        pre, cur = float("-inf"), 0
        cnt1, cnt2, cnt3 = 0, 0, 0
        i = 0
        while i < len(nums):
            cnt = 0
            cur = nums[i]
            while i < len(nums) and cur == nums[i]:
                cnt += 1
                i += 1

            if cur != pre + 1:
                if cnt1 != 0 or cnt2 != 0:
                    return False
                cnt1, cnt2, cnt3 = cnt, 0, 0
            else:
                if cnt < cnt1 + cnt2:
                    return False
                cnt1, cnt2, cnt3 = max(0, cnt - (cnt1 + cnt2 + cnt3)), \
                                   cnt1, \
                                   cnt2 + min(cnt3, cnt - (cnt1 + cnt2))
            pre = cur
        return cnt1 == 0 and cnt2 == 0

Python:

def isPossible(self, nums):
        left = collections.Counter(nums)
        end = collections.Counter()
        for i in nums:
            if not left[i]: continue
            left[i] -= 1
            if end[i - 1] > 0:
                end[i - 1] -= 1
                end[i] += 1
            elif left[i + 1] and left[i + 2]:
                left[i + 1] -= 1
                left[i + 2] -= 1
                end[i + 2] += 1
            else:
                return False
        return True　　

C++:

class Solution {
public:
	bool isPossible(vector& nums)
	{
		unordered_map, std::greater>> backs;

		// Keep track of the number of sequences with size < 3
		int need_more = 0;

		for (int num : nums)
		{
			if (! backs[num - 1].empty())
			{	// There exists a sequence that ends in num-1
				// Append 'num' to this sequence
				// Remove the existing sequence
				// Add a new sequence ending in 'num' with size incremented by 1 
				int count = backs[num - 1].top();
				backs[num - 1].pop();
				backs[num].push(++count);

				if (count == 3)
					need_more--;
			}
			else
			{	// There is no sequence that ends in num-1
				// Create a new sequence with size 1 that ends with 'num'
				backs[num].push(1);
				need_more++;
			}
		}
		return need_more == 0;
	}
};

C++:　　

class Solution {
public:
    bool isPossible(vector& nums) {
        unordered_map freq, need;
        for (int num : nums) ++freq[num];
        for (int num : nums) {
            if (freq[num] == 0) continue;
            else if (need[num] > 0) {
                --need[num];
                ++need[num + 1];
            } else if (freq[num + 1] > 0 && freq[num + 2] > 0) {
                --freq[num + 1];
                --freq[num + 2];
                ++need[num + 3];
            } else return false;
            --freq[num];
        }
        return true;
    }
};

　　

　　

 

 

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转载于:https://www.cnblogs.com/lightwindy/p/9808139.html