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;} |
原文：http://www.cnblogs.com/lautsie/p/3525973.html