[LeetCode] 659. Split Array into Consecutive Subsequences 将数组分割成连续子序列
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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]
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<Integer, Integer> 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<int>& nums) { unordered_map<int, priority_queue<int, vector<int>, std::greater<int>>> 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<int>& nums) { unordered_map<int, int> 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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