本文已参与「新人创作礼」活动，一起开启掘金创作之路。
原文链接：blog.csdn.net/roufoo/arti…
Zigzag Iterator II Follow up Zigzag Iterator: What if you are given k 1d vectors? How well can your code be extended to such cases? 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”.
Example Example1
Input: k = 3 vecs = [ [1,2,3], [4,5,6,7], [8,9], ] Output: [1,4,8,2,5,9,3,6,7] Example2
Input: k = 3 vecs = [ [1,1,1] [2,2,2] [3,3,3] ] Output: [1,2,3,1,2,3,1,2,3]
解法1：用轮询方式。 代码如下：
class ZigzagIterator2 {
public:
/*
* @param vecs: a list of 1d vectors
*/
ZigzagIterator2(vector<vector<int>>& vecs) {
vecsSize = vecs.size();
pollIndex = -1;
iters.resize(vecsSize);
iterEnds.resize(vecsSize);
for (int i = 0; i < vecsSize; ++i) {
iters[i] = vecs[i].begin();
iterEnds[i] = vecs[i].end();
}
}
/*
* @return: An integer
*/
int next() {
if (hasNext()) {
for (int i = pollIndex + 1; i < vecsSize; ++i) {
if (iters[i] != iterEnds[i]) {
pollIndex = i;
return *(iters[i]++);
}
}
for (int i = 0; i <= pollIndex; ++i) {
if (iters[i] != iterEnds[i]) {
pollIndex = i;
return *(iters[i]++);
}
}
}
}
/*
* @return: True if has next
*/
bool hasNext() {
for (int i = max(0, pollIndex); i < vecsSize; ++i) {
if (iters[i] != iterEnds[i]) return true;
}
for (int i = 0; i <= pollIndex; ++i) {
if (iters[i] != iterEnds[i]) return true;
}
return false;
}
private:
vector<vector<int>::iterator> iters, iterEnds;
//vector<vector<int>>::iterator pollIter, pollIterEnd;
int pollIndex, vecsSize;
};
/**
* Your ZigzagIterator2 object will be instantiated and called as such:
* ZigzagIterator2 solution(vecs);
* while (solution.hasNext()) result.push_back(solution.next());
* Ouptut result
*/
