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)
 */

扩展阅读

哈夫曼编码和游程编码