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Zigzag Iterator Given two 1d vectors, implement an iterator to return their elements alternately.
Example Example1
Input: v1 = [1, 2] and v2 = [3, 4, 5, 6] Output: [1, 3, 2, 4, 5, 6] Explanation: By calling next repeatedly until hasNext returns false, the order of elements returned by next should be: [1, 3, 2, 4, 5, 6]. Example2
Input: v1 = [1, 1, 1, 1] and v2 = [3, 4, 5, 6] Output: [1, 3, 1, 4, 1, 5, 1, 6]
解法1：借鉴那个Flattern 2D vector题的思路。
代码如下：
class ZigzagIterator {
public:
/*
* @param v1: A 1d vector
* @param v2: A 1d vector
*/
ZigzagIterator(vector<int>& v1, vector<int>& v2) {
iter1 = v1.begin();
v1End = v1.end();
iter2 = v2.begin();
v2End = v2.end();
iter = v1.end();
}
/*
* @return: An integer
*/
int next() {
if (hasNext()) {
if (iter1 != v1End && iter2 != v2End) {
if (iter == v1End) { //initial state
iter = iter1++;
} else if (iter == iter1 - 1) {
iter = iter2++;
} else {
iter = iter1++;
}
} else if (iter1 != v1End) {
iter = iter1++;
} else { // if (iter2 != v2End) {
iter = iter2++;
}
}
return *iter;
}
/*
* @return: True if has next
*/
bool hasNext() {
return !(iter1 == v1End && iter2 == v2End);
}
private:
vector<int>::iterator iter, iter1, iter2, v1End, v2End;
};
/**
* Your ZigzagIterator object will be instantiated and called as such:
* ZigzagIterator solution(v1, v2);
* while (solution.hasNext()) result.push_back(solution.next());
* Ouptut result
*/
