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  • leetcode-p2

    • leetcode261 Graph Valid Tree-zh
    • leetcode262 Trips and Users-zh
    • leetcode263 ugly-number-zh
    • leetcode264 Ugly Number II-zh
    • leetcode265 Paint House II-zh
    • leetcode266 Palindrome Permutation-zh
    • leetcode267 Palindrome Permutation II-zh
    • leetcode268 Missing Number-zh
    • leetcode269 Alien Dictionary-zh
    • leetcode270 Closest Binary Search Tree Value-zh
    • leetcode271 Encode and Decode Strings-zh
    • leetcode272 Closest Binary Search Tree Value II-zh
    • leetcode273 Integer to English Words-zh
    • leetcode274 H-Index-zh
    • leetcode275 H-Index II-zh
    • leetcode276 Paint Fence-zh
    • leetcode277 Find the Celebrity-zh
    • leetcode278 First Bad Version-zh
    • leetcode279 Perfect Squares-zh
    • leetcode280 Wiggle Sort-zh
    • leetcode281 Zigzag Iterator-zh
      • 题目描述
      • 解法
        • Python3
        • Java
        • ...
    • leetcode282 Expression Add Operators-zh
    • leetcode283 Move Zeroes-zh
    • leetcode284 Peeking Iterator-zh
    • leetcode285 Inorder Successor in BST-zh
    • leetcode286 Walls and Gates-zh
    • leetcode287 Find the Duplicate Number-zh
    • leetcode288 Unique Word Abbreviation-zh
    • leetcode289 Game of Life-zh
    • leetcode290 Word Pattern-zh
  • 算法
  • leetcode-p2
geekzl.com
2021-07-19

leetcode281 Zigzag Iterator-zh

# 281. 锯齿迭代器 (opens new window)

English Version (opens new window)

# 题目描述

给出两个一维的向量,请你实现一个迭代器,交替返回它们中间的元素。

示例:

输入:
v1 = [1,2]
v2 = [3,4,5,6] 

输出: [1,3,2,4,5,6]

解析: 通过连续调用 next 函数直到 hasNext 函数返回 false,
     next 函数返回值的次序应依次为: [1,3,2,4,5,6]。

拓展:假如给你 k 个一维向量呢?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:
[1,2,3]
[4,5,6,7]
[8,9]

输出: [1,4,8,2,5,9,3,6,7].

# 解法

定义 vectors 列表保存输入的所有一维向量,indexes 表示 vectors 列表每一项当前所遍历到的下标位置,cur 表示当前遍历到的 vector 列表,而 size 表示 vectors 列表元素个数。具体实现参考以下代码实现。

# Python3

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())
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# Java

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();
 */
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# ...


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#leetcode
上次更新: 2021/10/30, 12:58:38
leetcode280 Wiggle Sort-zh
leetcode282 Expression Add Operators-zh

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