900. RLE Iterator RLE 迭代器
作者： 负雪明烛 id： fuxuemingzhu 个人博客： http://fuxuemingzhu.cn/
题目地址：https://leetcode.com/problems/rle-iterator/description/
题目描述：
Write an iterator that iterates through a run-length encoded sequence.
The iterator is initialized by RLEIterator(int[] A), where A is a run-length encoding of some sequence. More specifically, for all even i, A[i] tells us the number of times that the non-negative integer value A[i+1] is repeated in the sequence.
The iterator supports one function: next(int n), which exhausts the next n elements (n >= 1) and returns the last element exhausted in this way. If there is no element left to exhaust, next returns -1 instead.
For example, we start with A = [3,8,0,9,2,5], which is a run-length encoding of the sequence [8,8,8,5,5]. This is because the sequence can be read as "three eights, zero nines, two fives".
Example 1:
Input: ["RLEIterator","next","next","next","next"], [[[3,8,0,9,2,5]],[2],[1],[1],[2]]
Output: [null,8,8,5,-1]
Explanation:
RLEIterator is initialized with RLEIterator([3,8,0,9,2,5]).
This maps to the sequence [8,8,8,5,5].
RLEIterator.next is then called 4 times:
.next(2) exhausts 2 terms of the sequence, returning 8. The remaining sequence is now [8, 5, 5].
.next(1) exhausts 1 term of the sequence, returning 8. The remaining sequence is now [5, 5].
.next(1) exhausts 1 term of the sequence, returning 5. The remaining sequence is now [5].
.next(2) exhausts 2 terms, returning -1. This is because the first term exhausted was 5,
but the second term did not exist. Since the last term exhausted does not exist, we return -1.
Note:
- 0 <= A.length <= 1000 A.length is an even integer. 0 <= A[i] <= 10^9 There are at most 1000 calls to RLEIterator.next(int n) per test case. Each call to RLEIterator.next(int n) will have 1 <= n <= 10^9.
题目大意
给出了一个数组，这个数组第偶数个位置表示的是其后面的那个数字出现的次数。让我们设计一个函数，能找出后面的第n个数字，如果后面没有数字就返回-1.这个运算的过程中要把偶数位置出现的次数给统计进去。
解题方法
这个设计确实类似于python的next()函数，估计next()是这个题的原型吧。
看了下A的规模是1000，那么对时间复杂度的要求并没有那么严格。其实这个题可以直接使用O(N)的时间复杂度求解。
使用index指向我们现在统计到的数组A的位置，这个位置只指向偶数位置。然后每次next()调用同时维护数组A和index。直接遍历就好，如果A[index]的个数>n，就往后去找，直至找到对应的位置，这个时候的位置就代表我们的找到了要弹出的数字。遍历的过程中一定要更新n，也要更新每个位置出现次数。这样就相当于n把原来位置的数字给消耗掉了。
平均时间复杂度是O(n)，空间复杂度是O(1)。
代码如下：
class RLEIterator(object):
def __init__(self, A):
"""
:type A: List[int]
"""
self.A = A
self.index = 0
def next(self, n):
"""
:type n: int
:rtype: int
"""
while self.index < len(self.A) and self.A[self.index] < n:
n -= self.A[self.index]
self.index += 2
if self.index >= len(self.A):
return -1
self.A[self.index] -= n
return self.A[self.index + 1]
# Your RLEIterator object will be instantiated and called as such:
# obj = RLEIterator(A)
# param_1 = obj.next(n)
二刷的时候使用C++，思路和上面大致相同。如果当前位置的数字大于等于剩余的n，那么把当前数字-n；如果当前位置的数字小于n，那么必须向后寻找了，需要更新n并且向后走，每次走两步。
如果上面的循环结束的时候，pos已经走到了数组外边，直接返回-1，否则返回pos位置对应的下一个位置。
C++代码如下：
class RLEIterator {
public:
RLEIterator(vector<int> A) : A_(A), pos(0){
}
int next(int n) {
while (pos < A_.size() && n > 0) {
if (A_[pos] >= n) {
A_[pos] -= n;
n = 0;
} else {
n -= A_[pos];
A_[pos] = 0;
pos += 2;
}
}
if (pos >= A_.size() - 1) return -1;
return A_[pos + 1];
}
private:
vector<int> A_;
int pos;
};
/**
* Your RLEIterator object will be instantiated and called as such:
* RLEIterator obj = new RLEIterator(A);
* int param_1 = obj.next(n);
*/
参考资料：
https://leetcode.com/problems/rle-iterator/discuss/168294/Java-Straightforward-Solution-O(n)-time-O(1)-space
日期
2018 年 9 月 18 日 —— 铭记这一天 2019 年 2 月 26 日 —— 二月就要完了