281. 锯齿迭代器 🔒

题目描述

给出两个整数向量 v1 和 v2，请你实现一个迭代器，交替返回它们的元素。

实现 ZigzagIterator 类：

ZigzagIterator(List<int> v1, List<int> v2) 用两个向量 v1 和 v2 初始化对象。
boolean hasNext() 如果迭代器还有元素返回 true，否则返回 false。
int next() 返回迭代器的当前元素并将迭代器移动到下一个元素。

示例 1:

输入：v1 = [1,2], v2 = [3,4,5,6]
输出：[1,3,2,4,5,6]
解释：通过重复调用 next 直到 hasNext 返回 false，那么 next 返回的元素的顺序应该是：[1,3,2,4,5,6]。

示例 2:

输入：v1 = [1], v2 = []
输出：[1]

示例 3:

输入：v1 = [], v2 = [1]
输出：[1]

拓展：假如给你 k 个向量呢？你的代码在这种情况下的扩展性又会如何呢?

拓展声明：
“锯齿” 顺序对于 k > 2 的情况定义可能会有些歧义。所以，假如你觉得 “锯齿” 这个表述不妥，也可以认为这是一种 “循环”。例如：

输入：v1 = [1,2,3], v2 = [4,5,6,7], v3 = [8,9]
输出：[1,4,8,2,5,9,3,6,7]

解法

方法一

Python3JavaRust

class ZigzagIterator:
    def __init__(self, v1: List[int], v2: List[int]):
        self.cur = 0
        self.size = 2
        self.indexes = [0] * self.size
        self.vectors = [v1, v2]

    def next(self) -> int:
        vector = self.vectors[self.cur]
        index = self.indexes[self.cur]
        res = vector[index]
        self.indexes[self.cur] = index + 1
        self.cur = (self.cur + 1) % self.size
        return res

    def hasNext(self) -> bool:
        start = self.cur
        while self.indexes[self.cur] == len(self.vectors[self.cur]):
            self.cur = (self.cur + 1) % self.size
            if self.cur == start:
                return False
        return True


# Your ZigzagIterator object will be instantiated and called as such:
# i, v = ZigzagIterator(v1, v2), []
# while i.hasNext(): v.append(i.next())

public class ZigzagIterator {
    private int cur;
    private int size;
    private List<Integer> indexes = new ArrayList<>();
    private List<List<Integer>> vectors = new ArrayList<>();

    public ZigzagIterator(List<Integer> v1, List<Integer> v2) {
        cur = 0;
        size = 2;
        indexes.add(0);
        indexes.add(0);
        vectors.add(v1);
        vectors.add(v2);
    }

    public int next() {
        List<Integer> vector = vectors.get(cur);
        int index = indexes.get(cur);
        int res = vector.get(index);
        indexes.set(cur, index + 1);
        cur = (cur + 1) % size;
        return res;
    }

    public boolean hasNext() {
        int start = cur;
        while (indexes.get(cur) == vectors.get(cur).size()) {
            cur = (cur + 1) % size;
            if (start == cur) {
                return false;
            }
        }
        return true;
    }
}

/**
 * Your ZigzagIterator object will be instantiated and called as such:
 * ZigzagIterator i = new ZigzagIterator(v1, v2);
 * while (i.hasNext()) v[f()] = i.next();
 */

struct ZigzagIterator {
    v1: Vec<i32>,
    v2: Vec<i32>,
    /// `false` represents `v1`, `true` represents `v2`
    flag: bool,
}

impl ZigzagIterator {
    fn new(v1: Vec<i32>, v2: Vec<i32>) -> Self {
        Self {
            v1,
            v2,
            // Initially beginning with `v1`
            flag: false,
        }
    }

    fn next(&mut self) -> i32 {
        if !self.flag {
            // v1
            if self.v1.is_empty() && !self.v2.is_empty() {
                self.flag = true;
                let ret = self.v2.remove(0);
                return ret;
            }
            if self.v2.is_empty() {
                let ret = self.v1.remove(0);
                return ret;
            }
            let ret = self.v1.remove(0);
            self.flag = true;
            return ret;
        } else {
            // v2
            if self.v2.is_empty() && !self.v1.is_empty() {
                self.flag = false;
                let ret = self.v1.remove(0);
                return ret;
            }
            if self.v1.is_empty() {
                let ret = self.v2.remove(0);
                return ret;
            }
            let ret = self.v2.remove(0);
            self.flag = false;
            return ret;
        }
    }

    fn has_next(&self) -> bool {
        !self.v1.is_empty() || !self.v2.is_empty()
    }
}

281. 锯齿迭代器 🔒

题目描述

解法

方法一

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