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。
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; }