將一個數組拆分爲幾個至少三個元素的遞增子序列

Split Array into Consecutive Subsequences

問題：

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]

解決：

①  給定的整數數組按照升序排列（可能包含重複項），須要將它們分紅幾個子序列，每一個子序列至少包含3個連續的整數。 返回是否能夠作出這樣的分割。

1. 遍歷數組，獲得數組中全部元素的出現次數；
2. 再次遍歷數組，而且對於每一個元素，要麼查看它是否能夠附加到先前構建的連續序列，要麼是否能夠是新的連續序列的開始。 若是兩個都不是真的，那咱們就返回假。

class Solution { //108ms
    public boolean isPossible(int[] nums) {
        Map<Integer,Integer> frequenceMap = new HashMap<>();
        Map<Integer,Integer> appendMap = new HashMap<>();
        for (int n : nums){
            frequenceMap.put(n,frequenceMap.getOrDefault(n,0) + 1);
        }
        for (int n : nums){
            if (frequenceMap.get(n) == 0) continue;
            else if (appendMap.getOrDefault(n,0) > 0){
                appendMap.put(n,appendMap.get(n) - 1);
                appendMap.put(n + 1,appendMap.getOrDefault(n + 1,0) + 1);
            }else if (frequenceMap.getOrDefault(n + 1,0) > 0 && frequenceMap.getOrDefault(n + 2,0) > 0){
                frequenceMap.put(n + 1,frequenceMap.get(n + 1) - 1);
                frequenceMap.put(n + 2,frequenceMap.get(n + 2) - 1);
                appendMap.put(n + 3,appendMap.getOrDefault(n + 3,0) + 1);
            }else {
                return false;
            }
            frequenceMap.put(n,frequenceMap.get(n) - 1);
        }
        return true;
    }
} 

② 在discuss中看到的解法。。。

class Solution {//16ms
    public boolean isPossible(int[] nums) {
        int pre = Integer.MIN_VALUE, p1 = 0, p2 = 0, p3 = 0;
        int cur = 0, cnt = 0, c1 = 0, c2 = 0, c3 = 0;
        for (int i = 0; i < nums.length; pre = cur, p1 = c1, p2 = c2, p3 = c3) {
            for (cur = nums[i], cnt = 0; i < nums.length && cur == nums[i]; cnt++, i++);
            if (cur != pre + 1) {
                if (p1 != 0 || p2 != 0) return false;
                c1 = cnt; c2 = 0; c3 = 0;
            } else {
                if (cnt < p1 + p2) return false;
                c1 = Math.max(0, cnt - (p1 + p2 + p3));
                c2 = p1;
                c3 = p2 + Math.min(p3, cnt - (p1 + p2));
            }
        }
        return p1 == 0 && p2 == 0;
    }
}