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  • 281. Zigzag Iterator

    题目:

    Given two 1d vectors, implement an iterator to return their elements alternately.

    For example, given two 1d vectors:

    v1 = [1, 2]
    v2 = [3, 4, 5, 6]
    

    By calling next repeatedly until hasNext returns false, the order of elements returned by next should be: [1, 3, 2, 4, 5, 6].

    Follow up: What if you are given k 1d vectors? How well can your code be extended to such cases?

    Clarification for the follow up question - Update (2015-09-18):
    The "Zigzag" order is not clearly defined and is ambiguous for k > 2 cases. If "Zigzag" does not look right to you, replace "Zigzag" with "Cyclic". For example, given the following input:

    [1,2,3]
    [4,5,6,7]
    [8,9]
    

    It should return [1,4,8,2,5,9,3,6,7].

    链接: http://leetcode.com/problems/zigzag-iterator/

    题解:

    Zigzag Iterator。最简单的就是用两个index分别遍历两个list,然后用一个boolean变量来控制何时遍历哪一个list。 好像我这么做是有问题的,应该用Interator<>来做。

    Time Complexity - O(n), Space Complexity - O(1)

    public class ZigzagIterator {
        private int length;
        private int index1 = 0;
        private int index2 = 0;
        private List<Integer> list1;
        private List<Integer> list2;
        private boolean useList2;
        
        public ZigzagIterator(List<Integer> v1, List<Integer> v2) {
            this.length = v1.size() + v2.size();
            list1 = v1;
            list2 = v2;
            if(list1.size() == 0) {
                useList2 = true;
            }
        }
    
        public int next() {
            if(!useList2) {
                if(index2 < list2.size()) {
                    useList2 = true;    
                }
                return list1.get(index1++);
            } else {
                if(index1 < list1.size()) {
                    useList2 = false;
                }
                return list2.get(index2++);
            }
       
        }
    
        public boolean hasNext() {
            return (index1 + index2) < length;
        }
    }
    
    /**
     * Your ZigzagIterator object will be instantiated and called as such:
     * ZigzagIterator i = new ZigzagIterator(v1, v2);
     * while (i.hasNext()) v[f()] = i.next();
     */

    Reference:

    https://leetcode.com/discuss/57961/o-n-time-%26-o-1-space-java-solution

    https://leetcode.com/discuss/63037/simple-java-solution-for-k-vector

    https://leetcode.com/discuss/58012/short-java-o-1-space

    https://leetcode.com/discuss/71857/clean-java-solution-works-for-k-lists

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  • 原文地址:https://www.cnblogs.com/yrbbest/p/5035622.html
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