这一题也简单，唯一有意思的地方是提炼了一个函数用来做数组索引去重前进。
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int
forward(vector<int> &arr, int
i) {
while
(i+1 < arr.size() && arr[i] == arr[i+1]) i++;
i++;
return
i;
}vector<int> arrayUnion(vector<int> &a, vector<int> &b) {
vector<int> ans;
int
i = 0;
int
j = 0;
while
(i < a.size() && j < b.size()) {
if
(a[i] == b[j]) {
ans.push_back(a[i]);
i = forward(a, i);
j = forward(b, j);
} else
if
(a[i] < b[j]) {
ans.push_back(a[i]);
i = forward(a, i);
} else
{
ans.push_back(b[j]);
j = forward(b, j);
}
}
while
(i < a.size()) {
ans.push_back(a[i]);
i = forward(a, i);
}
while
(j < b.size()) {
ans.push_back(b[j]);
j = forward(b, j);
}
return
ans;
}vector<int> arrayIntersect(vector<int> &a, vector<int> &b) {
vector<int> ans;
int
i = 0;
int
j = 0;
while
(i < a.size() && j < b.size()) {
if
(a[i] == b[j]) {
ans.push_back(a[i]);
i = forward(a, i);
j = forward(b, j);
} else
if
(a[i] < b[j]) {
i = forward(a, i);
} else
{
j = forward(b, j);
}
}
return
ans;
} |