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.
思路
这是一个游程编码的典型题目。
该算法分为两个部分,一个是初始化,一个是调用next(n)
.
我们需要做的就是初始化的时候,记住这个A。 然后每次调用next(n)
的时候只需要
判断n是否大于Ai
-
如果大于A[i], 那就说明不够,我们移除数组前两项,更新n,重复1
-
如果小于A[i], 则说明够了,更新A[i]
这样做,我们每次都要更新A,还有一种做法就是不更新A,而是伪更新
,即用一个变量记录,当前访问到的数组位置。
很多时候我们需要原始的,那么就必须这种放了,我的解法就是这种方法。
关键点解析
代码
/* * @lc app=leetcode id=900 lang=javascript * * [900] RLE Iterator * * https://leetcode.com/problems/rle-iterator/description/ * * algorithms * Medium (49.03%) * Total Accepted: 11.6K * Total Submissions: 23.5K * Testcase Example: '["RLEIterator","next","next","next","next"]\n[[[3,8,0,9,2,5]],[2],[1],[1],[2]]' * * 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. * * */ /** * @param {number[]} A */ var RLEIterator = function(A) { this.A = A; this.current = 0; }; /** * @param {number} n * @return {number} */ RLEIterator.prototype.next = function(n) { const A = this.A; while(this.current < A.length && A[this.current] < n){ n = n - A[this.current]; this.current += 2; } if(this.current >= A.length){ return -1; } A[this.current] = A[this.current] - n; // 更新Count return A[this.current + 1]; // 返回element }; /** * Your RLEIterator object will be instantiated and called as such: * var obj = new RLEIterator(A) * var param_1 = obj.next(n) */
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